Learning Objectives
After studying this chapter you should be able to:
- Identify a random experiment, an outcome and a trial, and give everyday examples of randomness.
- Explain probability as a measurement of likelihood on a scale from \(0\) (impossible) to \(1\) (certain).
- Use the words of probability precisely: impossible, unlikely, equally likely, more likely and certain.
- Distinguish experimental probability (from evidence) from theoretical probability (from equally likely reasoning).
- Find the experimental probability of an outcome from a frequency table, and write any probability as a fraction, decimal or percentage.
- List the sample space \(S\) of an experiment without duplicating or missing outcomes, and state \(n(S)\).
- Recognise an event as a subset of the sample space, and find the probability of an event when outcomes are equally likely.
- Build tree diagrams for two-stage experiments and read probabilities from them, distinguishing with replacement from without replacement.
- Explain the Law of Large Numbers and the gambler’s fallacy, and why a fair coin or die has no memory.
- Connect probability to data, sampling and everyday decisions, and interpret what a probability such as \(0.16\) or \(40\%\) actually promises.
1. It’s All About Maybe: Uncertainty Is Everywhere
Think about how often your day is described with words like maybe, perhaps, probably, likely, unlikely, chance, odds or fifty-fifty. “It might rain today.” “Our school could win the trophy.” “There is a chance you get picked for the lucky draw.” We use these words all the time, yet they are vague — what exactly does “probably” mean? The mathematics in this chapter replaces such fuzzy words with precise numbers. The moment we say “the probability is \(0.75\)” or “the chance is \(\frac{3}{4}\)”, everyone agrees on exactly how likely an event is.
Key Idea
Random event: an event for which we know all the possible outcomes, but we do not know in advance which one will definitely occur. There is an element of chance or randomness involved every time such an event takes place.
Here are three questions you might genuinely ask or be asked:
- Is it going to rain today?
- Will our school win the inter-school hockey match tomorrow?
- Will I be chosen in the monthly lucky draw to perform at the school assembly? (All the names of the students are written on slips of paper and one slip is randomly selected.)
Each of these is a random event: we know the possible outcomes — it will rain or it will not; our team will win, draw or lose; exactly one student will be chosen — but we cannot say with certainty which one will happen.
Think and Reflect
Ask a friend to predict the outcome of a one-rupee coin you are about to toss. Do you see that your friend could guess heads or tails but could not know for certain?
That is randomness! All the possible results are known, but each individual try is unpredictable.
We respond to such questions with words such as impossible or certain, or with phrases such as less likely, more likely or equally likely. These choices rest on evidence. For example, one friend might say, “The sun is shining brightly, so it’s unlikely to rain today”, while another might observe, “It’s very hot, which makes me think it could rain later.” Both are predicting rainfall by interpreting the present weather conditions in their own way. This is the subjective probability given to the event of today’s rainfall.
Probability is the area of mathematics that studies randomness and how likely a specific outcome is to happen in a random situation. One key feature of our increasingly complicated society is that we must deal with questions that have no fixed answer but rather one or more possibilities for the answer. Understanding how to estimate the probability of events objectively is therefore crucial in many aspects of life.
2. What Is Probability?
Definition
Probability: a type of measurement, similar to how we measure length, area or volume. But instead of measuring physical quantities, probability measures the likelihood of an event — it helps us express how confident or certain we are that a particular event will occur.
Because probability is a measurement, it comes with a number. Suppose the probability that your school wins the inter-school hockey match is \(0.75\). That means there is a \(75\%\) chance your school will win, so it is more likely than not to win. If the probability is \(0.5\), there is a \(50\%\) chance, meaning it is equally likely that your school wins or loses. If the probability is \(0\), winning is impossible (for example, winning without playing). If the probability is \(1\), your school is certain to win.
Example 1 — Reading a Probability
A weather report says, “There is a 70% chance of rain this afternoon.” What does this number tell us?
Solution: The report is giving us a probability of \(0.7\) for rain. It does not say it will rain or that it will not rain; it says that, comparing many afternoons with the same conditions, rain is expected on 70 out of every 100 of them. Rain is more likely than no rain, but not certain.
Remember
A probability lets us talk about how likely an outcome is. It never lets us predict a single random result exactly.
3. What Is Randomness?
Definition
Randomness: a situation or action (like the tossing of a coin or the rolling of a die) where you cannot predict exactly what will happen. Although you may know all the possible outcomes, you cannot say which one will definitely occur.
- Tossing a coin: you know it could be heads or tails, but you cannot be sure which one will come up in a single toss.
- Rolling a die: you know the possible outcomes are \(1, 2, 3, 4, 5\) or \(6\), but you do not know which number will appear on a particular roll.
Key Term
Random experiment: something you can repeat (like tossing a coin), where every time you do it the result might be different, and you cannot know the outcome in advance. These observations are also called experiments or trials in probability.
The lucky draw mentioned earlier is also a random experiment: the outcome cannot be predicted in advance, and each student has an equal chance of being selected. All you can say is what could happen, not what will happen.
Think and Reflect
Such unpredictability can be useful sometimes! In a cricket match, the fact that a coin is tossed to decide which team will bat first is considered to be a fair method. Can you explain why?
Because nobody can influence the outcome of the toss. Heads and tails are equally likely, so the decision is made by chance alone and neither team can be unfairly favoured.
Have you wondered what makes an event like rain random? Rain depends on many complex factors in the atmosphere — temperature, humidity, wind patterns, pressure — and it is so sensitive to these factors that it is impossible to predict it with total certainty. That is why we cannot know with absolute certainty whether it will rain on a specific day. In the same way that the exact timing and location of rainfall cannot be predicted perfectly, we can still estimate the likelihood of rain in different places based on patterns and data.
For example, when you toss a coin, the probability of heads is \(\frac{1}{2}\), and the probability of tails is \(\frac{1}{2}\), because each is equally likely. In a random experiment every outcome has a chance to occur, but you can only determine the likelihood — never the exact result.
4. The Probability Scale
Definition
Probability scale: probability is measured on a scale from \(0\) to \(1\). An event that is impossible has probability \(0\). An event that is certain has probability \(1\). The probabilities of most events lie strictly between \(0\) and \(1\): the closer a probability is to \(1\), the more likely the event.
Just like a number line, this scale lets us compare how likely different events are. A probability of \(0.75\) is the same as a \(75\%\) chance; a probability of \(0.5\) is a \(50\%\) chance, an even chance; a probability of \(0.8\) means the event is more likely than one with probability \(0.4\), and so on.
The probability of picking a purple card from this deck moves smoothly along the scale as the number of purple cards changes: from less likely, to even chance, to more likely. This scale helps us understand and compare how likely different events are — just like using a number line!
| Event | What it means |
|---|---|
| Getting a number greater than 6 on a die | Impossible: dice only have numbers from 1 to 6. |
| Rolling a 3 on a standard die | Less likely: not impossible, because one face of the die is 3. |
| Flipping a coin and getting heads | Even chance: heads or tails are equally likely. |
| Drawing any number from 2 to 10 from a deck of 52 cards | More likely: there are 36 cards in a deck of 52 with numbers from 2 to 10. |
| Choosing a red sweet from a bag of all red sweets | Certain: all the sweets are red. |
Exercise Set 7.1 (with reasoning)
Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons.
(i) The next Monday will come after Sunday.
(ii) It will snow in Mumbai in July.
(iii) An elephant will walk through your classroom today.
(iv) You will greet at least one friend at school tomorrow.
(i) Certain (1) — the days of the week always repeat in this order, without exception. (ii) Impossible (0) — Mumbai has a hot tropical climate in July, and it never snows there. (iii) Impossible (0) — an elephant cannot possibly walk through your classroom today. (iv) More likely to certain — unless you are absent, you will almost certainly cross paths with at least one friend; the exact label depends on your situation, which is exactly why we give reasons.
5. Words We Use in Probability
Before we measure, we need a precise vocabulary. Each word below corresponds to a position on the probability scale from \(0\) to \(1\).
Impossible
A situation that can never happen. Example: getting a number greater than \(6\) on a die. Probability \(0\).
Unlikely / Less likely
Could happen, but usually does not. Example: rolling a \(3\) on a standard die. Probability between \(0\) and \(\frac{1}{2}\).
Equally likely / Even chance
Two outcomes that have the same chance, like heads and tails of a fair coin. Each has probability \(\frac{1}{2}\).
More likely
Happens more often than not, but is not guaranteed. Example: drawing a number from 2 to 10 out of a full deck. Probability between \(\frac{1}{2}\) and \(1\).
Certain
A situation that always happens. Example: choosing a red sweet from a bag of all red sweets. Probability \(1\).
Remember
The words are only useful when we are clear about the situation. “It will rain tomorrow” and “it will not rain tomorrow” is not a 50-50 pair — rain depends on many factors, and the chances are usually uneven.
6. Measuring Probability Objectively
Subjective probability depends on a person’s belief and judgement. Different people can honestly give different probabilities for the same event. To communicate well and compare fairly, we need probabilities that anyone can reproduce from the same evidence. There are two main ways to estimate the probability of an event objectively:
| Method | Idea | Resulting probability |
|---|---|---|
| Using evidence from experience | Collect data by performing an experiment many times, or by analysing statistical data from past observations. | Experimental probability (the relative frequency of the event). |
| Using theoretical methods | Assume all possible outcomes are equally likely (when there is no reason to believe any one outcome is more likely than another), and reason about favourable outcomes. | Theoretical probability |
Both methods give a number between \(0\) and \(1\). Later in this chapter you will see that in perfectly fair situations the two numbers tend to come very close when the experiment is repeated many times.
7. Experimental Probability
One way to obtain an objective estimate is to perform an experiment. In an experiment, the event is a result of the experiment, called an outcome. The set of all possible outcomes is called the sample space. The sample space is often listed within brackets, separated by commas.
Example 2 — Outcomes and Sample Spaces
Experiment: a coin is tossed.
Possible outcomes: Heads (H) or Tails (T).
Sample space: \(S = \{H, T\}\).
Experiment: a die is rolled.
Possible outcomes: the top side shows \(1, 2, 3, 4, 5\) or \(6\) dots.
Sample space: \(S = \{1, 2, 3, 4, 5, 6\}\).
Formula
Experimental probability of an event
\[ \text{Experimental probability} = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}} \]
Example 3 — Experimental Probability from an Experiment
Suppose you roll a die 50 times and it lands on a 4 exactly 8 times.
Solution: the experimental probability of rolling a 4 is
\[ \frac{8}{50} = 0.16 = 16% \]
We also say that the relative frequency of rolling a 4 is \(0.16\). Relative frequency helps us understand probability based on actual data rather than theoretical predictions; it is especially useful in statistics and data analysis when we work with observed outcomes.
Did You Know?
The game of Snakes and Ladders originates in ancient India. It evolved from a traditional dice game called Jñān-Chaupad, which was used as an educational tool. It aimed to convey moral lessons, with each ladder symbolising a virtue and each snake representing a vice; through gameplay it illustrated the consequences of good and bad behaviour. Various versions of Jñān-Chaupad have been found in different parts of India.
8. Experimental-Probability Activity: Toss a Coin 20 Times
Let us carry out the experiment from Exercise Set 7.2, Question 3. To explain the method, here is one illustrative set of 20 tosses of a coin (record your own results in the same way):
| Toss | Result | Toss | Result |
|---|---|---|---|
| 1 | H | 11 | T |
| 2 | H | 12 | H |
| 3 | T | 13 | T |
| 4 | H | 14 | H |
| 5 | T | 15 | H |
| 6 | H | 16 | H |
| 7 | H | 17 | T |
| 8 | T | 18 | H |
| 9 | H | 19 | H |
| 10 | T | 20 | T |
In this illustrative set, heads appears \(12\) times and tails \(8\) times. So:
- Experimental probability of heads \(= \frac{12}{20} = 0.6 = 60%\).
- Experimental probability of tails \(= \frac{8}{20} = 0.4 = 40%\).
Careful
This is one way the experiment can turn out — another student performing the same 20 tosses could easily get \(9\) heads and \(11\) tails, or even \(13\) heads and \(7\) tails. The experimental probability is exactly what your data gives; it is not guaranteed to equal \(\frac{1}{2}\). As we will see, more trials bring us closer to \(\frac{1}{2}\).
Activity
Toss a paper cup into the air 100 times. After each toss, record whether the cup lands on its bottom, upside down on its top, or on its side.
When you finish, assign probabilities to the three outcomes using experimental probability: the number of times each position occurred, divided by 100. For example, if the cup lands on its side 61 times, the experimental probability of landing on its side is \(\frac{61}{100} = 0.61\).
9. Relative Frequency
Definitions
Frequency: the number of times an outcome occurred, such as \(12\) heads in 20 tosses.
Relative frequency: the frequency divided by the total number of trials. \[ \text{Relative frequency} = \frac{\text{Frequency of the outcome}}{\text{Total number of trials}} \]
Experimental probability is exactly the relative frequency of the event in the data collected.
Example 4 — Relative Frequency from a School Survey
In a physical-education class, a student attempts 60 free throws and makes 42 of them.
Solution: the relative frequency of a successful free throw is \(\frac{42}{60} = 0.7\). We may estimate that the probability of this student making a free throw is about \(0.7\), or \(70%\), based on the evidence.
Relative frequency is the bridge between data and probability. Whenever we have a frequency table — from an experiment, a survey or a record of past observations — we can turn each frequency into a probability-style estimate. You will use this idea repeatedly, because real-world decisions (how many mangoes to buy for the school, how reliable a product is, how often a bus is late) are made from such evidence.
10. Theoretical Probability
The second way to measure likelihood uses reasoning instead of data. Theoretical probability studies the likelihood of an event on the assumption that all possible outcomes are equally likely — what we expect to happen in an ideal, perfectly fair situation. In this scenario no experiment or data is required.
Definition
Theoretical probability of an event \(A\):
\[ P(A) = \frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} \]
and it is usually denoted \(P(\text{Event})\) or \(P(\text{Outcome})\).
Condition
The formula above is valid only when every possible outcome is equally likely. We must always check this assumption before applying it — you will see in Section 18 why a careless assumption can give a wrong answer.
Example 5 — Theoretical Probability of Rolling a 4
If you roll a standard 6-sided die, what is the theoretical probability of getting a 4?
Solution: a fair die makes all six faces equally likely.
- Number of favourable outcomes \(= 1\) (only the number 4).
- Number of all possible outcomes \(= 6\) (the numbers 1 through 6).
\[ P(\text{rolling a } 4) = \frac{1}{6} = 0.1666\ldots \approx 0.167 \text{ or } 16.7% \]
Example 6 — Picking a Letter from PROBABILITY
A letter is picked at random from the word PROBABILITY. What is the probability of picking the letter B?
Solution: assuming each letter card is equally likely to be picked:
- Number of favourable outcomes \(= 2\) (there are two B’s in PROBABILITY).
- Number of all possible outcomes \(= 11\) (the number of letters in the word).
\[ P(\text{picking the letter } B) = \frac{2}{11} = 0.1818\ldots \approx 0.182 \text{ or } 18.2% \]
11. Probability in Different Forms
A single probability can be written as a fraction, a decimal or a percentage; they are all the same number. To convert a fraction to a decimal, divide the numerator by the denominator. To convert a decimal to a percentage, multiply by \(100%\).
| Fraction | Decimal | Percentage | Meaning |
|---|---|---|---|
| \( \dfrac{1}{4} \) | \(0.25\) | \(25%\) | One in four chances |
| \( \dfrac{1}{2} \) | \(0.5\) | \(50%\) | Even chance |
| \( \dfrac{3}{4} \) | \(0.75\) | \(75%\) | Three in four chances |
| \( \dfrac{2}{5} \) | \(0.4\) | \(40%\) | Two in five chances |
| \( \dfrac{1}{6} \) | \(0.1666\ldots \approx 0.167\) | \(\approx 16.7%\) | One in six chances |
Example 7 — Writing a Probability Three Ways
A pack has 10 pencils, of which 4 are red. One pencil is drawn at random. Express \(P(\text{red})\) as a fraction, a decimal and a percentage.
Solution:
\[ P(\text{red}) = \frac{4}{10} = \frac{2}{5} \]
As a decimal: \(\dfrac{2}{5} = 2 \div 5 = 0.4\). As a percentage: \(0.4 \times 100% = 40%\).
12. Analysing Statistical Data Using Probability
Evidence from statistical data is used extensively in business — for marketing, forecasting sales and insurance — as well as in science and social-science research. The method is the same relative-frequency idea: turn collected data into a probability estimate.
Example 8 — The Favourite-Fruit Game
Suppose you anonymously collect information about the favourite fruit of 50 students in your class. The results are: 20 students like mango, 15 like apples, 10 like bananas, and 5 like grapes.
Let us play a game! Suppose we randomly pick one student from the class and try to guess their favourite fruit. What is the probability that the student’s favourite fruit is mango?
Solution: statistical probability is based on collected data, so a reasonable estimate is
\[ P(\text{mango is the favourite fruit}) = \frac{\text{Number of students who like mango}}{\text{Total number of students}} = \frac{20}{50} = 0.4 \]
So there is a \(0.4\) or \(40%\) chance that a randomly chosen student in the class likes mango.
Now suppose you want to buy fruits for the entire school according to the students’ favourite fruits. It is usually impractical to collect data from all students. The school (the population) has 1500 students, but we collected evidence from a sample of 50 students in one class. Based on our estimate of \(0.4\), we should buy approximately
\[ 0.4 \times 1500 = 600 \text{ mangoes} \]
while the remaining fruits would be apples, bananas and grapes. If we want to be more confident about our estimate, we would choose a larger sample (for example, asking 100 students) and make sure it is more representative, such as including students from different classes or grades. This process is called sampling in statistics.
13. Sampling: Evidence from a Sample
Definitions
Population: the whole group we want to know about (for example, all 1500 students of the school).
Sample: the smaller group from whom we actually collect data (for example, the 50 students in one class).
Sampling: the process of choosing a sample and using it to estimate facts about the population.
What we have learned so far is just the beginning of sampling. When collecting data through samples there are other important points to think about:
- How big your sample should be — a larger sample generally gives more accurate results.
- How representative it is — a truly useful sample reflects the variety of the whole population (for example, students from different classes and grades, not just one class).
- Fairness and bias — the sample should not be biased in a way that favours one answer (for example, asking only mango-lovers would overestimate the demand for mangoes).
Remember
A sample is only useful if it is big enough and fair enough to represent the population. That is why a survey from one classroom is informative, but a survey from many classrooms is much more trustworthy.
14. The Law of Large Numbers
Let us summarise the two measurements and how they are related:
- Experimental probability is based on actual data collected from trials, not on assumptions.
- Theoretical probability relies on the assumption of equally likely outcomes (a perfectly fair situation) and does not use experimental data.
Even in perfectly fair situations, experimental probability can differ from theoretical probability — especially when the number of trials is small. As the number of trials increases, the experimental probability tends to get closer to the theoretical probability. This is known as the Law of Large Numbers.
Key Idea
Law of Large Numbers: the more times we repeat an experiment, the closer the experimental probability tends to come to the theoretical probability. For example, the experimental probability of rolling a 3 will stabilise around \(\frac{1}{6}\) as we roll many more times.
Notice what the law does not say. It does not say that after some number of trials the experimental probability will suddenly equal the theoretical probability. It says the experimental probability tends to get closer as trials increase — the zigzag becomes a gentle wobble around the true value.
Think and Reflect
If I have rolled a 4 on a die 8 times in succession, the probability of rolling a 4 again is still only \(\approx 0.16\) (assuming the die is fair). Probability does not tell you what will happen next, but predicts what will happen in the long run.
15. Independent Repeated Trials
Definition
Independent trials: trials are independent when the outcome of one trial does not change the probability of the outcomes of the next trial. The result of one coin toss or die roll in no way changes the result of the next one.
Experiments such as tossing a coin twice, rolling a die three times, or spinning a wheel of numbers repeatedly are called multi-step experiments because they involve a series of independent trials. Independence is a key idea: it lets us combine probabilities from separate steps, which is exactly what tree diagrams (Section 24) will do.
- If you toss a fair coin and get heads, the chance of heads on the next toss is still \(\frac{1}{2}\).
- If you roll a die and get a 6, the chance of a 6 on the next roll is still \(\frac{1}{6}\).
Careful
Tossing a coin twice and then adding the results is independent only if the first toss is not allowed to influence the second — which is true for a fair coin that is tossed freely. Whenever something is drawn without being replaced, the trials stop being independent (see Section 26).
16. The Gambler’s Fallacy
Many people think that if something random happens many times in a row (like flipping a coin and getting heads six times), then the opposite outcome (tails) is more likely to happen next. But actually, the coin does not keep track of what happened before. Every time you flip, the chance of heads or tails stays exactly the same — like starting afresh. The coin has no memory, so past flips do not change what happens next. This common misunderstanding is called the Gambler’s Fallacy.
Example 9 — Six Heads in a Row
You are flipping a fair coin and it comes up heads six times in a row. You might feel that on the next flip you are sure to get a tail, because tails haven’t come up for a while.
Solution: the probability of getting tails on the next flip is still \(50%\), i.e. \(\frac{1}{2}\), exactly the same as before. The coin has no memory of past flips!
Example 10 — Three Sixes in Snakes and Ladders
Let us say you are playing Snakes and Ladders and you are rolling a fair 6-sided die to move. You have just rolled the die three times in a row, and each time you got a 6. Now you think: “I have already rolled three 6s — there is no way I will get a 6 again on the next roll!”
Solution: this thinking is the Gambler’s Fallacy — because each roll of the die is an independent event. The probability of rolling a 6 is always
\[ \frac{1}{6} \approx 0.166 \text{ or } 16.6% \]
It does not change based on what happened in previous turns.
Key Idea
In Snakes and Ladders, as in many games of chance, each roll of the dice is independent. The Gambler’s Fallacy tricks people into believing there is a pattern, when in fact such randomness has no memory.
17. Fair and Unbiased
Key Terms
Fair: when we speak of a coin, we assume it to be “fair” — that is, it is symmetrical so that there is no reason for it to come down more often on one side than the other.
Unbiased: this property of the coin — no side is favoured — is called being unbiased.
Random toss: by this phrase we mean that the coin is allowed to fall freely without any bias or interference.
All the theoretical probabilities we compute in this chapter — \(\frac{1}{2}\) for heads, \(\frac{1}{6}\) for each face of a die, \(\frac{1}{36}\) for a particular pair of dice — rest on the assumption that the objects are fair and the trials are unbiased.
Remember
Fairness is an assumption we use when reasoning. A real coin might be slightly heavier on one side, and a real die might have slightly uneven faces. That is why experimental probability exists — it measures what actually happens, fairness or not.
18. Elements of Probability: Sample Spaces and Events
Every random experiment has three companions: the sample space, the event, and the probability. Everything we do from here on is a movement through this chain:
The sample space is the “menu” of everything that can happen; an event is a choice from that menu (perhaps several items); and the probability measures how often the event happens in the long run.
19. Sample Space
Definition
Sample space: the list of all possible outcomes of a random experiment, denoted by \(S\). Each possible outcome is called an element of the sample space. The number of elements in the sample space is called the sample size, denoted by \(n(S)\).
A valid sample space must obey three rules:
- The sample space \(S\) must include every possible outcome.
- No outcome should be listed more than once.
- The sample size is \(n(S)\), the count of elements.
Examples 11 – 12 — Sample Spaces of Simple Situations
11. The sample space for whether it rains tomorrow could be \(S = \{\text{Rain, No Rain}\}\), with sample size \(n(S) = 2\).
12. The sample space for the possible results of a team in a match could be \(S = \{\text{Win, Lose, Draw}\}\), with sample size \(n(S) = 3\).
Think and Reflect
When we used the sample space \(\{\text{Rain, No Rain}\}\), we focused only on whether it will rain or not. However, if we want to include different amounts of rainfall like drizzle, light rain or heavy rain, we need to expand the sample space to \(\{\text{No Rain, Drizzle, Light Rain, Heavy Rain}\}\) so that it better matches the level of detail required for the question.
It is important to ensure the sample space is detailed enough to suit the specific problem being studied. A coarse sample space is fine for a coarse question; a fine question needs a fine sample space.
20. Sample Spaces for Different Experiments
Let us write the sample space for several experiments, always applying the three rules of Section 19: complete, no repeats, and counted.
| Experiment | Sample space \(S\) | Sample size \(n(S)\) |
|---|---|---|
| Tossing a fair coin once | \(S = \{H, T\}\) | \(2\) |
| Rolling a standard 6-sided die | \(S = \{1, 2, 3, 4, 5, 6\}\) | \(6\) |
| Tossing two coins together | \(S = \{HH, HT, TH, TT\}\) | \(4\) |
| Rolling a die and tossing a coin together | \(S = \{1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T\}\) | \(12\) |
| Tossing a coin and drawing a card numbered 1 to 6 | \(S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}\) | \(12\) |
| Choosing a random integer between \(-5\) and \(+5\) | \(S = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}\) | \(11\) |
| Drawing one ball from a box of 5 green and 7 red balls | \(S = \{\text{Green, Red}\}\) | \(2\) |
For the two-coin experiment, the sample space is easiest to trust when we list it step by step:
| Coin 1 | Coin 2 | Outcome |
|---|---|---|
| H | H | HH |
| H | T | HT |
| T | H | TH |
| T | T | TT |
Why 12, not 6?
For a die and a coin, the most common mistake is to count only the six die faces. But each die face can come with either coin side, so the size is \(6 \times 2 = 12\). When one experiment has \(a\) outcomes and another has \(b\) outcomes, together they have \(a \times b\) combined outcomes.
Exercise Set 7.3 — Snacks at the Village Fair
In a village fair, there are 3 popular snacks available: Samosa, Pakora and Bhaji. For drinks, villagers can choose either Chai or Lassi.
(i) List the sample space of all possible snack-and-drink combinations a person could choose at the fair.
(ii) List the event “Selecting Samosa as a snack.”
(i) \(3 \times 2 = 6\) combinations: {Samosa-Chai, Samosa-Lassi, Pakora-Chai, Pakora-Lassi, Bhaji-Chai, Bhaji-Lassi}. (ii) The event is “Samosa with Chai” and “Samosa with Lassi”.
21. Events
Definition
Event: any single possible result or combination of results that might happen when you perform a random action. It is like choosing particular outcomes from all the things that could possibly occur. Formally, an event is a subset of a sample space.
In symbols we write an event as \(E\) and a pair of braces around its outcomes, just like the sample space. Let us see three examples from the textbook.
Examples 13 – 15 — Sample Spaces and Events
13. Tossing two coins. Sample space \(S = \{HH, HT, TH, TT\}\). Event “at least one coin shows Head” is \(E = \{HH, HT, TH\}\).
14. Rolling a 6-sided die. Sample space \(S = \{1, 2, 3, 4, 5, 6\}\). Event “the number rolled is greater than 4” is \(E = \{5, 6\}\).
15. Picking fruit from a basket. Sample space \(S = \{\text{Apple, Banana, Orange}\}\). Event “picking a fruit that is yellow” is \(E = \{\text{Banana}\}\).
Careful
An event with one outcome is still correctly written with braces, e.g. \(E = \{5\}\). Every event — even a single outcome — lives inside the sample space \(E \subseteq S\).
22. Types of Events
Simple event
Exactly one outcome, such as rolling \(4\): \(E = \{4\}\).
Multi-outcome event
Two or more outcomes, such as rolling an even number: \(E = \{2, 4, 6\}\).
Impossible event
No outcome in the sample space satisfies it, so \(E = \{\}\) and \(P(E) = 0\). Example: rolling a number greater than 6 on a die.
Certain event
Every outcome satisfies it, so \(E = S\) and \(P(E) = 1\). Example: rolling a number less than 7 on a die.
Example 16 — Building Events by Hand
An event has probability \(1\)? The event “the sum of two dice is between 2 and 12” is certain — every pair of dice sums to at least 2 and at most 12 — so its probability is \(1\). An event has probability \(0\)? The event “the sum of two dice is 13” is impossible, so its probability is \(0\).
Solution: impossible events have no favourable outcomes (probability \(0\)); certain events contain the whole sample space (probability \(1\)); everything else lies strictly between.
23. Tree Diagrams
Definition
Tree diagram: a visual representation used to list all possible outcomes of a multi-step experiment. A multi-step experiment involves a series of independent trials — for example, tossing a coin two times or rolling a die three times. Each branch of the tree represents a possible outcome, and branches split to show different paths for subsequent events.
Tree diagrams are useful for:
- Visualising multi-step experiments — each path from start to end represents one complete outcome.
- Listing all outcomes of a sample space — no repetition, no omission.
Example 17 — Toss a Fair Coin Two Times
From a single point, draw a line to each possible outcome of the first toss. From each of these outcomes, draw two lines to each possible outcome of the second toss. Do you see that the tree diagram shows 4 possible outcomes?
Hence the sample space is \(S = \{HH, HT, TH, TT\}\), with \(n(S) = 4\).
24. Probabilities on a Tree
The theoretical probability of each outcome is written on the branch representing that outcome. For a fair coin, each first-step branch carries \(\frac{1}{2}\), and so does each second-step branch. Along a path, the probabilities multiply: the path HH has probability
\[ P(HH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25 = 25% \]
Similarly \(P(HT) = \frac{1}{4}\), \(P(TH) = \frac{1}{4}\) and \(P(TT) = \frac{1}{4}\), and the four probabilities on the leaves add up to \(1\) — a nice check that we have not missed or invented any outcome.
Think and Reflect
Can you calculate the probability of getting one head and one tail?
The event “one head and one tail” is \(E = \{HT, TH\}\), so \(P(E) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\). Just two of the four equally likely paths are favourable.
Example 18 — Three Coins, Exactly Two Heads
Three fair coins are tossed together. What is the probability of getting exactly two heads?
Solution: the tree for three coins has \(2 \times 2 \times 2 = 8\) leaves, all equally likely. Exactly two heads occur in \(\{HHT, HTH, THH\}\), which is 3 paths:
\[ P(\text{exactly two heads}) = \frac{3}{8} \]
25. Drawing With and Without Replacement
Key Terms
With replacement: after the first item is drawn, it is put back before the next draw. The situation is unchanged for the second draw, so the trials are independent and every stage has the same number of outcomes.
Without replacement: the first item is set aside and not returned. For the second draw there is one fewer item, the chances shift, and the trials are no longer independent.
Example 19 — Balls Numbered 1 to 4
A box contains 4 balls numbered 1 to 4.
(i) With replacement: a ball is drawn and its number recorded, then the ball is returned, and a second ball is drawn and recorded. Each draw has 4 possible numbers, so the sample space has \(4 \times 4 = 16\) pairs, from \(11\) up to \(44\).
(ii) Without replacement: a ball is drawn and recorded, then — without replacing the first ball — a second ball is drawn. The first draw has 4 possibilities and the second only 3 (the first ball is gone), so the sample space has \(4 \times 3 = 12\) pairs, and a pair like \(11\) or \(22\) cannot occur.
(iii) The sizes of the two sample spaces are \(16\) and \(12\).
Remember
Watch how the question is worded: “and put it back” means with replacement; “drawn and laid aside” means without replacement. The tree diagram shows all possibilities clearly — because the second-stage branches lose one option in the without-replacement case.
26. Tree Diagram Applications
Example 20 — Fruit Baskets (Exercise Set 7.4, Q1)
There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.
Solution:
(i)–(ii) The tree has two stages. From A the fruit can be an apple or an orange; from B it can be a banana or a mango. The sample space is
\[ S = \{\text{Apple-Banana, Apple-Mango, Orange-Banana, Orange-Mango}\} \]
with \(n(S) = 4\).
(iii) The event “one apple and one banana” is exactly the single pair \(\{\text{Apple-Banana}\}\). Treating the four pairs as equally likely outcomes,
\[ P(\text{apple and banana}) = \frac{1}{4} \]
State Your Model
Here we are counting fruit types, so the sample space has four pairs. If instead we counted each individual orange as a separate item, basket A would give apple probability \(\frac{1}{3}\) and orange probability \(\frac{2}{3}\), and the answer would change. This is why we always write down the sample space — it records exactly the model we are using.
Example 21 — Pens With Replacement (Exercise Set 7.4, Q2)
A box contains 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same. What is the probability that both of you pick pens of the same colour?
Solution: there are \(3 + 4 + 2 = 9\) pens. The first pick has probabilities \(\frac{3}{9}\) (red), \(\frac{4}{9}\) (black) and \(\frac{2}{9}\) (green). Because the first pen is put back, the second pick has exactly the same probabilities — the tree has 9 leaves.
Both pick the same colour in three ways:
\[ \begin{aligned} P(\text{both red}) &= \frac{3}{9} \times \frac{3}{9} = \frac{9}{81} \\ P(\text{both black}) &= \frac{4}{9} \times \frac{4}{9} = \frac{16}{81} \\ P(\text{both green}) &= \frac{2}{9} \times \frac{2}{9} = \frac{4}{81} \end{aligned} \]
\[ P(\text{same colour}) = \frac{9}{81} + \frac{16}{81} + \frac{4}{81} = \frac{29}{81} \]
Example 22 — Red and Blue Balls Without Replacement
A basket contains 4 red and 5 blue balls. One ball is drawn and laid aside, and then a second ball is drawn. Draw a tree diagram and find: (i) the probability of a red ball then a blue ball; (ii) the probability of two blue balls.
Solution: the first draw has \(P(R) = \frac{4}{9}\) and \(P(B) = \frac{5}{9}\). Since the first ball is laid aside, the second-stage branches take denominators of \(8\): if the first ball was red, \(P(R) = \frac{3}{8}\) and \(P(B) = \frac{5}{8}\); if the first ball was blue, \(P(R) = \frac{4}{8}\) and \(P(B) = \frac{4}{8}\). Therefore:
\[ P(\text{red then blue}) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18} \]
\[ P(\text{two blue}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18} \]
27. Key Definitions
| Term | Meaning |
|---|---|
| Probability | A measurement of the likelihood of an event, on a scale from \(0\) to \(1\). |
| Random event | An event whose possible outcomes are known but whose exact outcome is not. |
| Random experiment / trial | A repeatable action whose result may differ each time and cannot be predicted in advance. |
| Outcome | A single possible result of an experiment. |
| Sample space \(S\) | The list of all possible outcomes of an experiment; every outcome once, none omitted. |
| Sample size \(n(S)\) | The number of elements in the sample space. |
| Event \(E\) | A subset of the sample space; any single outcome or combination of outcomes. |
| Equally likely outcomes | Outcomes that have the same chance of occurring, with no reason to prefer one. |
| Fair / unbiased | An object or process that favours no outcome, e.g. a symmetric coin. |
| Experimental probability | The relative frequency of an event in collected data: \(\dfrac{\text{event count}}{\text{total trials}}\). |
| Relative frequency | The frequency of an outcome divided by the total number of trials. |
| Theoretical probability | \(\dfrac{\text{favourable outcomes}}{\text{possible outcomes}}\), valid when all outcomes are equally likely. |
| Impossible event | An event with no favourable outcome; probability \(0\). |
| Certain event | An event that always occurs, equal to the whole sample space; probability \(1\). |
| Population and sample | The whole group being studied and the smaller group whose data we actually collect. |
| Sampling | Choosing a representative sample and using it to estimate facts about the population. |
| Tree diagram | A branching diagram that lists all outcomes of a multi-step experiment. |
| Independent trials | Trials whose outcomes do not influence each other. |
| Law of Large Numbers | As trials increase, experimental probability tends to approach theoretical probability. |
| Gambler’s Fallacy | The mistaken belief that an unusual run of outcomes must soon be balanced out. |
28. Formula Bank
All the Formulas in One Place
- Every probability lies on the scale: \(0 \le P(E) \le 1\).
- Experimental probability (from evidence): \[ \frac{\text{Number of times the event occurred}}{\text{Total number of trials}} \]
- Relative frequency: \[ \frac{\text{Frequency of the outcome}}{\text{Total number of trials}} \]
- Theoretical probability (only when outcomes are equally likely): \[ P(A) = \frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} \]
- Combined experiments: \(a\) outcomes followed by \(b\) outcomes give \(a \times b\) combined outcomes.
- Along a tree path: multiply the branch probabilities.
- Two ways to get the same event: add the path probabilities, e.g. \(P(HT) + P(TH)\).
- Complimentary event: \(P(\text{not } E) = 1 - P(E)\).
29. Important Concepts
Probability is a measurement
Like length or area, but it measures likelihood. It always comes with a number between 0 and 1.
The scale 0 to 1
0 impossible, 1 certain, \(\frac{1}{2}\) even chance, 0.75 “more likely than not”.
Two objective approaches
Evidence from experience gives experimental probability; equally likely reasoning gives theoretical probability.
Experimental = relative frequency
Count divided by total trials. Grows stable as trials increase.
Theoretical needs equally likely
Check the assumption first. The formula fails when outcomes are not equally likely.
Sample space rules
Complete, no duplicates, then count it: \(n(S)\).
Events are subsets
\(E \subseteq S\). A single outcome is an event too.
Trees for multi-steps
Each path is one outcome. Multiply along branches, add across paths.
No memory
Fair coins and dice are independent; the Gambler’s Fallacy is just a myth.
Law of Large Numbers
More trials → experimental probability tends to the theoretical value.
30. More Solved Examples
Example 23 — Sweets Sample (Exercise Set 7.2, Q1)
A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts 10 red, 8 green, 7 yellow and 5 blue sweets.
(i) Probability that a randomly picked sweet from the sample is green:
\[ P(\text{green}) = \frac{8}{30} = \frac{4}{15} \approx 0.267 \approx 26.7% \]
(ii) If there are 600 sweets in the large bag, estimate the number likely to be yellow:
\[ \frac{7}{30} \times 600 = 140 \text{ sweets} \]
Example 24 — Favourite-Club Survey (Exercise Set 7.2, Q2)
A random sample of 40 students is asked about their favourite club: 14 prefer Science, 11 Arts, 9 Sports and 6 Debate. The whole school has 800 students.
(i) Probability that a randomly chosen student from the sample prefers the Arts Club:
\[ P(\text{Arts}) = \frac{11}{40} = 0.275 \]
(ii) Estimated number preferring the Sports Club in the whole school:
\[ \frac{9}{40} \times 800 = 180 \text{ students} \]
Example 25 — Twelve Rolls, Three Threes (Exercise Set 7.2, Q6)
You roll a 6-sided die 12 times and get a 3 three times.
(i) Experimental probability of rolling a 3: \[ \frac{3}{12} = \frac{1}{4} = 0.25 \]
(ii) Theoretical probability of rolling a 3 (fair die): \[ \frac{1}{6} \approx 0.167 \]
(iii) They differ because 12 trials is a small sample and chance jiggles the results. As you roll 60, 600 or 6000 times, the experimental probability tends to settle toward \(\frac{1}{6}\) (the Law of Large Numbers).
Example 26 — Are the Outcomes Equally Likely?
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car: it starts or does not start — not equally likely; a well-maintained car starts far more often than it fails.
(ii) Tossing a fair coin once — equally likely.
(iii) Rolling a fair 6-sided die — equally likely.
(iv) Choosing a marble at random from a bag with 3 red and 7 blue marbles — not equally likely; blue has \(7/10\), red only \(3/10\).
(v) A baby is born: a boy or a girl — not exactly equally likely; birth records show a small, consistent excess of boys.
Example 27 — Tyre Company Records (End-of-Chapter, Q7)
A tyre company records the distance covered before replacement in 1000 cases: less than 4000 km — 20; 4001 to 9000 km — 210; 9001 to 14000 km — 325; more than 14000 km — 445.
(i) Less than 4000 km: \( \dfrac{20}{1000} = 0.02 \)
(ii) Between 4000 and 14000 km: \( \dfrac{210 + 325}{1000} = \dfrac{535}{1000} = 0.535 \)
(iii) More than 14000 km: \( \dfrac{445}{1000} = 0.445 \)
Example 28 — Spinner at a Game Booth (End-of-Chapter, Q9)
A game of chance consists of spinning an arrow that points to one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, all equally likely.
(i) Point at 8: \( \dfrac{1}{8} \)
(ii) Odd number: \( \dfrac{4}{8} = \dfrac{1}{2} \)
(iii) Number greater than 2: \( \dfrac{6}{8} = \dfrac{3}{4} \)
(iv) Number less than 9: \( \dfrac{8}{8} = 1 \) (certain)
(v) Multiple of 3: \( \dfrac{2}{8} = \dfrac{1}{4} \)
Example 29 — Sum of Two Dice Is Prime and Greater Than 5
Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
Solution: the largest possible sum is 12, so the primes greater than 5 are 7 and 11.
Sum 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) — 6 outcomes.
Sum 11: \((5,6),(6,5)\) — 2 outcomes.
Favourable \(= 6 + 2 = 8\), possible \(= 36\):
\[ P = \frac{8}{36} = \frac{2}{9} \]
Example 30 — Four-Digit Even Numbers
A four-digit number is formed using the digits 1, 2, 3 and 4, with no repetition. What is the probability that the number is even?
Solution: total arrangements \(= 4 \times 3 \times 2 \times 1 = 24\). For an even number the last digit must be 2 or 4 (2 choices), and the other three digits can be arranged in \(3 \times 2 \times 1 = 6\) ways.
Favourable \(= 2 \times 6 = 12\):
\[ P(\text{even}) = \frac{12}{24} = \frac{1}{2} \]
Example 31 — Guessing in a Multiple-Choice Test
A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?
Solution: choose which 2 questions are correct: 3 ways. For those two, the probability of guessing correctly is \(\left(\frac{1}{4}\right)^2\) and for the remaining question the probability of being wrong is \(\frac{3}{4}\).
\[ P(\text{exactly 2 correct}) = 3 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4} = 3 \times \frac{1}{16} \times \frac{3}{4} = \frac{9}{64} \]
Example 32 — Three Coins, First Head and Exactly Two Heads
Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
Solution: exactly two heads with the first coin a head gives the two paths HHT and HTH out of 8 equally likely paths:
\[ P = \frac{2}{8} = \frac{1}{4} \]
Example 33 — Balls of Different Colours
A bag contains 4 red, 3 green and 2 blue balls. Two balls are drawn without replacement. What is the probability that the two balls are of different colours?
Solution: total 9 balls. It is often easier to find the complementary event, “same colour”:
\[ \begin{aligned} P(\text{same}) &= \frac{4}{9}\cdot\frac{3}{8} + \frac{3}{9}\cdot\frac{2}{8} + \frac{2}{9}\cdot\frac{1}{8} \\ &= \frac{12}{72} + \frac{6}{72} + \frac{2}{72} = \frac{20}{72} \end{aligned} \]
\[ P(\text{different}) = 1 - \frac{20}{72} = \frac{52}{72} = \frac{13}{18} \]
Example 34 — Dye Drop on a Rectangular Region
Suppose you drop a dye at random on a rectangular region 2 m by 3 m. What is the probability that it will land inside a circle with a diameter of 1 m placed in the rectangle?
Solution: the circle has radius \(0.5\) m and area \(\pi(0.5)^2 = \frac{\pi}{4} \approx 0.785\text{ m}^2\). The rectangle has area \(2 \times 3 = 6\text{ m}^2\). Assuming every point is equally likely to be hit,
\[ P(\text{inside the circle}) = \frac{\pi/4}{6} = \frac{\pi}{24} \approx 0.131 \approx 13.1% \]
31. Common Real-Life Applications
- Weather forecasting: “70% chance of rain” is a probability estimate built from past weather patterns — used to plan outdoor events.
- Sports: the coin toss for the cricket match is fair because heads and tails are equally likely; teams estimate win chances from past performance.
- Surveys and market research: the favourite-fruit data (20 of 50 like mango) turns into an estimate of what an entire school will want to buy.
- Business and insurance: tyre companies record failure distances and use the relative frequencies to set warranties; insurers price premiums from accident probabilities.
- Games of chance: Snakes and Ladders, lucky draws, spinners and dice games all depend on the probabilities we have computed.
- Quality control: testing a small sample of products (like 30 sweets or 40 students) tells us about the whole batch — that is sampling in action.
- Transport and planning: the chance that a bus arrives within 5 minutes can be estimated from the relative frequency of past arrivals, helping travellers decide.
- Smart exam strategy: knowing that pure guessing on a 4-option MCQ gives exactly 2 of 3 correct with probability only \(\frac{9}{64} \approx 0.14\) discourages relying on luck.
32. Competency-Based Questions
Question A
Two students toss the same coin 20 times each. Student 1 records 12 heads and Student 2 records 9 heads. The theoretical probability of heads is \(\frac{1}{2}\). Explain why both students’ experimental probabilities (0.6 and 0.45) are correct and not contradictory.
Experimental probability is a description of the data actually collected; 20 trials is a small sample, so values around 0.5 are expected to wobble. Both have measured genuinely. Only as the number of tosses grows would we expect both results to settle near 0.5.
Question B
A class has one six-faced die. Zoe says the die is biased because she rolled a 3 only twice in 12 rolls. Using what you know about the Law of Large Numbers, write a short response to Zoe.
Two 3s in 12 rolls gives an experimental probability of 1/6 by coincidence, but even wildly different short runs (like 0 or 5 threes in 12 rolls) are possible with a fair die. To check for bias you need many more trials — hundreds or thousands — and compare the long-run relative frequency with 1/6.
Question C
A forecast says: “30% chance of rain tomorrow.” Neel says: “Since it is 70% likely not to rain, the forecast is telling me it will not rain.” Do you agree? Explain.
No. 70% not raining makes no-rain the more likely outcome, but rain is still possible 3 times out of 10. Probability measures likelihood, not certainty; the correct statement is “no rain is more likely, but rain is not ruled out.”
Question D
For a box with 5 green and 7 red balls, why is the sample space just \(S = \{\text{Green, Red}\}\) with \(n(S) = 2\), even though green is less likely than red?
The sample space lists the kinds of outcome, not their chances: a ball drawn is either green or red. Probability enters later — \(5/12\) and \(7/12\) — when we use the size of each colour. The size of the sample space and the likelihood of its elements are different ideas.
Question E
A friend’s tree diagram for a two-stage experiment has 9 leaves, but the probabilities on the leaves add to 1.1. What does that tell you?
One or more branch probabilities are wrong, or the paths are not being constructed independently. In a correct tree the leaf probabilities always sum to 1, so a total of 1.1 signals a mistake that must be found before trusting any answer.
Question F
“My coin landed heads 10 times in a row, so tails is due now.” Identify the flaw in this claim using the idea of independence.
This is the Gambler’s Fallacy. A fair coin has no memory: each toss is independent, so the probability of tails on the next toss is still exactly \(\frac{1}{2}\). Past heads do not “owe” us a tail.
33. Case-Based Questions
Case 1 — School Assembly Lucky Draw
A school has 240 students in Class 9. For the Republic Day performance, every student’s name is written on a slip, and one slip is drawn at random to perform a welcome dance.
(i) Write the sample space and find \(n(S)\).
(ii) Riya is the only student named Riya. What is the probability that she is chosen?
(iii) Exactly 108 of the 240 slips are girls’ names. What is the probability that a girl is chosen?
(i) \(S\) = the 240 names, \(n(S) = 240\). (ii) \(1/240\). (iii) \(108/240 = 9/20 = 0.45 = 45%\).
Case 2 — Fair-Day Spinner Game
At a school fair a spinner has the numbers 1 to 10, all equally likely. A child wins a prize if the arrow points to a number greater than 7.
(i) Find the probability of winning.
(ii) Find the probability of not winning.
(iii) The stall owner spins the arrow 300 times in a day. About how many times should she expect a win?
(i) Numbers 8, 9, 10: \(3/10 = 0.3\). (ii) \(1 - 0.3 = 0.7\). (iii) About \(0.3 \times 300 = 90\) wins.
Case 3 — Fruit-Basket Sampling in the Lunchroom
To plan a fruit giveaway, the school samples 60 students: 24 prefer apples, 18 bananas, 12 oranges and 6 pears. The school has 1200 students.
(i) Estimate the probability that a randomly chosen student prefers apples.
(ii) Estimate how many of the 1200 students prefer bananas.
(iii) Suggest one way to make the estimate more trustworthy.
(i) \(24/60 = 0.4 = 40%\). (ii) \(0.3 \times 1200 = 360\) (since \(18/60 = 0.3\)). (iii) Take a larger sample across many classes so it better represents the whole school, not just one lunchroom group.
Case 4 — Two-Stage Drawing With and Without Replacement
A box contains 2 red tokens and 3 blue tokens. Draw two tokens.
Scenario A: the first token is put back before the second draw. Scenario B: the first token is laid aside.
(i) Find the probability of drawing two red tokens in Scenario A.
(ii) Find the probability of drawing two red tokens in Scenario B.
(iii) Without calculating, say which scenario makes a second red more probable, and justify briefly.
(i) \((2/5)(2/5) = 4/25 = 0.16\). (ii) \((2/5)(1/4) = 2/20 = 0.1\). (iii) Scenario A: putting the first red back keeps the box unchanged, so the chance of a second red stays \(2/5\), whereas in Scenario B only 1 red remains among 4 tokens.
Case 5 — Running Records at the Sports Meet
A PE teacher records a student’s successful jumps: 52 of 80 jumps cleared the bar over the season.
(i) Find the relative frequency of a successful jump.
(ii) Estimate the probability that her next jump clears the bar.
(iii) Give a reason why this estimate could still be wrong at a competitive meet.
(i) \(52/80 = 0.65 = 65%\). (ii) Best estimate: about \(0.65\). (iii) Conditions change (tiredness, competition pressure or a different pit), so past records may not fully represent the new situation — evidence is never a promise.
Case 6 — Two Coins in a Board Game
A board game needs the result of tossing two coins. A player moves 2 steps if both coins show heads, 1 step if exactly one coin shows heads, and 0 steps if no coin shows heads.
(i) Write the sample space.
(ii) Find the probability of moving 2 steps, 1 step and 0 steps.
(iii) If the player tosses 80 times, about how many times will they move exactly 1 step?
(i) \(\{HH, HT, TH, TT\}\). (ii) \(P(2\text{ steps}) = 1/4\); \(P(1\text{ step}) = 2/4 = 1/2\); \(P(0\text{ steps}) = 1/4\). (iii) About \(0.5 \times 80 = 40\) times.
34. Common Mistakes
Each row below shows a wrong idea, the correct version, and why the mistake happens.
| Wrong | Correct | Why |
|---|---|---|
| After eight 4s in a row, a 4 is “due” and less likely. | The next roll still has \(P(4) = \frac{1}{6}\). | Each roll is independent; the die has no memory (Gambler’s Fallacy). |
| 12 heads in 20 tosses means \(P(\text{head}) = \frac{1}{2}\). | Experimental probability is \(\frac{12}{20} = 0.6\). | Experimental probability describes the data; the theoretical value stays \(\frac{1}{2}\). |
| \(P(\text{greater than 6 on a die}) = \frac{1}{6}\). | It is 0 — impossible. | The sample space \(\{1,2,3,4,5,6\}\) has no outcome greater than 6. |
| Die and coin together give 6 outcomes. | They give \(6 \times 2 = 12\). | Each die face pairs with each coin side. |
| Heads-count sample space for 3 coins is \(\{1, 2, 3\}\). | It is \(\{0, 1, 2, 3\}\). | 0 heads (all tails) and 3 heads must be included. |
| \(P(\text{not } E) = 1 + P(E)\). | \(P(\text{not } E) = 1 - P(E)\). | The event and its complement together cover probability 1. |
| Basket with 2 oranges and 1 apple gives \(P(\text{apple}) = \frac{1}{4}\). | In the fruit-type model yes; counted as objects it is \(\frac{1}{3}\). | Always state the model. The textbook exercise uses the four fruit-pair outcomes. |
| Writing a probability as \(1.2\) or \(120%\). | Probabilities stay between 0 and 1 (0% and 100%). | The probability scale ends at 1. |
| Saying “the frequency is \(\frac{8}{50}\)”. | Frequency is \(8\); relative frequency is \(\frac{8}{50}\). | Frequency is a count; relative frequency is a fraction. |
| \(P(\text{both heads in two tosses}) = \frac{1}{2}\). | \(P(HH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). | Two branches must each contribute; \(TH\) and \(HT\) are different outcomes. |
| Adding probabilities along a tree path. | Multiply along a path; add across paths. | Each step conditions the path; different paths are alternatives. |
| Same probabilities whether drawing with or without replacement. | Without replacement, the second denominator drops by 1. | The first item is gone, so outcomes shift. |
| \(P(E) = \dfrac{\text{favourable}}{\text{total}}\) for every situation. | Only when all outcomes are equally likely. | Otherwise the favourites are not equal chances. |
| Getting \(\frac{10}{6}\) as a probability. | Impossible — it exceeds 1. | Favourable outcomes can never outnumber possible ones. |
| \(0.167 = 0.167%\). | \(0.167 = 16.7%\). | To get a percentage, multiply the decimal by 100. |
| Ten heads in a row means tails is “owed”. | The next toss is still \(50%\) either way. | Randomness has no memory. |
Example 35 — Which List Is a Valid Sample Space?
Three coins are tossed and the number of heads is recorded. Which of these lists is the sample space, and why do the others fail?
(i) \(\{1, 2, 3\}\) (ii) \(\{0, 1, 2\}\) (iii) \(\{0, 1, 2, 3, 4\}\) (iv) \(\{0, 1, 2, 3\}\)
Solution: (iv) \(\{0, 1, 2, 3\}\) is correct. (i) leaves out 0; (ii) leaves out 3; (iii) includes 4, which cannot happen — a valid sample space must be complete and contain no impossible outcomes.
35. Exam Focus
Probability questions in examinations reward three habits: write the model, show the working, and give the final probability in the requested form. Practise these typical question styles:
| Skill | How it is tested | Strategy |
|---|---|---|
| Words and scale | Rank events as impossible, less likely, even chance, more likely, certain. | Anchor each word to a number on 0–1. |
| Experimental probability | Give a frequency table; find the probability of an outcome. | Event count ÷ total trials; simplify. |
| Theoretical probability | Fair coin, die, cards, letters, spinners. | Check equally likely, then favourable over possible. |
| Sample space | List \(S\) for combined experiments. | Complete, no repeats, then count \(n(S)\). |
| Events | Give an event as a subset and find its probability. | Write \(E = \{\ldots\}\), then count favourable. |
| Tree diagrams | Two-stage experiments, with and without replacement. | Label branch probabilities; multiply along paths. |
| Sampling | Estimate population counts from a sample. | Relative frequency × population size. |
| Reasoning | Explain independence, the Gambler’s Fallacy, the Law of Large Numbers. | Use the key ideas and a short example. |
Before you submit, check
- Is the probability between 0 and 1 (or 0% and 100%)?
- Did I state “equally likely” for theoretical probability?
- Did I use the right denominators (with or without replacement)?
- Does the answer match the required form (fraction, decimal or percentage)?
36. Quick Revision
Probability
A measurement of likelihood between 0 and 1: \(0 \le P(E) \le 1\).
The scale
0 impossible · less likely · even chance \(\frac{1}{2}\) · more likely · 1 certain.
Experimental probability
\(\dfrac{\text{times the event occurred}}{\text{total trials}}\).
Relative frequency
\(\dfrac{\text{frequency of the outcome}}{\text{total trials}}\).
Theoretical probability
\(\dfrac{\text{favourable}}{\text{possible}}\), only when outcomes are equally likely.
Sample space
All outcomes listed once; \(n(S)\) counts them.
Event
\(E \subseteq S\). Probability of an event \(=\) number of favourable elements over \(n(S)\) when equally likely.
Complement
\(P(\text{not } E) = 1 - P(E)\).
Tree diagrams
Paths list outcomes; multiply along branches, add across paths.
Replacement
With replacement: independent, same denominators. Without: denominators shrink.
No memory
The Gambler’s Fallacy is false: fair coins and dice forget the past.
Law of Large Numbers
More trials: experimental probability tends to the theoretical value.
Sampling
Use a fair, representative sample to estimate facts about the whole population.
37. Self-Practice Problems (SP1 – SP32)
- A fair coin is tossed once. Find the probability of heads.
- A fair die is rolled once. Find the probability of rolling a 5.
- A fair die is rolled once. Find the probability of rolling an odd number.
- Ten identical cards numbered 1 to 10 are placed in a box; one is drawn at random. Find the probability of drawing an even number.
- For the same cards as SP4, find the probability of drawing a number greater than 7.
- A spinner has numbers 1 to 8, all equally likely. Find the probability of landing on a prime number.
- A letter is chosen at random from the word STATISTICS. Find the probability of getting S.
- Two fair coins are tossed together. Find the probability of two heads.
- Two fair coins are tossed together. Find the probability of exactly one head.
- Two fair coins are tossed together. Find the probability of at least one tail.
- A bag has 5 red and 3 blue balls; one ball is drawn at random. Find \(P(\text{red})\).
- For the same bag as SP11, find \(P(\text{not red})\).
- A die is rolled and a coin is tossed together. How many outcomes are in the sample space?
- Three fair coins are tossed. Find the probability of exactly one head.
- Three fair coins are tossed. Find the probability of at least one head.
- A die is rolled 30 times and a 6 appears 4 times. Find the experimental probability of rolling a 6.
- Over how many of 300 fair rolls would you expect a 3 to appear?
- In a survey of 200 students, 44 prefer cycling. Estimate the probability that a randomly chosen student prefers cycling.
- A school has 900 students; a random sample of 60 shows 15 students like History. Estimate how many students in the school like History.
- A box has 3 red and 4 black pens. Two pens are drawn without replacement. Use a tree diagram to find \(P(\text{red then black})\).
- A box has 2 red and 3 blue pens. Two pens are drawn with replacement. Find the probability that both draws give the same colour.
- Letters of the word PEACE are put on cards. One card is drawn. Find \(P(E)\) and \(P(\text{not } E)\).
- An integer is chosen at random from \(-5\) to \(5\). Find the probability that it is negative.
- An integer is chosen at random from \(-5\) to \(5\). Find the probability that it is 0.
- Two dice are rolled. Find the probability that the sum is 5.
- Two dice are rolled. Find the probability that the sum is greater than 10.
- Two dice are rolled. Find the probability of a double (both faces equal).
- A spinner labelled 1, 2, 3, 4 is spun twice. Find the probability that the sum is 5.
- A test has 5 multiple-choice questions with 4 options each and one correct answer each. A student guesses every question. Find the probability of getting all 5 correct.
- A three-digit number is formed from the digits 1, 2, 3 without repetition. Find the probability that it is even.
- A class has 40 students, of whom 16 are boys. One student is chosen at random. Find the probability that a boy is chosen.
- On one bus route, 3 of 60 departures were more than 5 minutes late. Estimate the probability that a departure is more than 5 minutes late.
38. Self-Practice Answers
| Question | Answer (with working) |
|---|---|
| SP1 | \(\dfrac{1}{2} = 0.5 = 50%\) |
| SP2 | \(\dfrac{1}{6} \approx 0.167 \approx 16.7%\) |
| SP3 | Odd faces 1, 3, 5: \(\dfrac{3}{6} = \dfrac{1}{2}\) |
| SP4 | Even cards 2, 4, 6, 8, 10: \(\dfrac{5}{10} = \dfrac{1}{2}\) |
| SP5 | 8, 9, 10: \(\dfrac{3}{10}\) |
| SP6 | Primes 2, 3, 5, 7: \(\dfrac{4}{8} = \dfrac{1}{2}\) |
| SP7 | STATISTICS has 10 letters, S appears 3 times: \(\dfrac{3}{10}\) |
| SP8 | \(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\) |
| SP9 | HT or TH: \(\dfrac{2}{4} = \dfrac{1}{2}\) |
| SP10 | Not TT: \(1 - \dfrac{1}{4} = \dfrac{3}{4}\) |
| SP11 | \(\dfrac{5}{8}\) |
| SP12 | \(\dfrac{3}{8}\) (or \(1 - \frac{5}{8}\)) |
| SP13 | \(6 \times 2 = 12\) |
| SP14 | HTT, THT, TTH: \(\dfrac{3}{8}\) |
| SP15 | Not TTT: \(1 - \dfrac{1}{8} = \dfrac{7}{8}\) |
| SP16 | \(\dfrac{4}{30} = \dfrac{2}{15} \approx 0.133\) |
| SP17 | \(\dfrac{1}{6} \times 300 = 50\) times |
| SP18 | \(\dfrac{44}{200} = \dfrac{11}{50} = 0.22 = 22%\) |
| SP19 | \(\dfrac{15}{60} \times 900 = 225\) students |
| SP20 | \(\dfrac{3}{7} \times \dfrac{4}{6} = \dfrac{12}{42} = \dfrac{2}{7}\) |
| SP21 | \(\left(\frac{2}{5}\right)^2 + \left(\frac{3}{5}\right)^2 = \frac{4}{25} + \frac{9}{25} = \frac{13}{25}\) |
| SP22 | \(P(E) = \dfrac{2}{5}\); \(P(\text{not } E) = \dfrac{3}{5}\) |
| SP23 | \(-5\) to \(-1\) is 5 integers of 11: \(\dfrac{5}{11}\) |
| SP24 | \(\dfrac{1}{11}\) |
| SP25 | (1,4),(2,3),(3,2),(4,1): \(\dfrac{4}{36} = \dfrac{1}{9}\) |
| SP26 | Sum 11 (2 ways) + sum 12 (1 way): \(\dfrac{3}{36} = \dfrac{1}{12}\) |
| SP27 | (1,1)…(6,6): \(\dfrac{6}{36} = \dfrac{1}{6}\) |
| SP28 | Total \(4 \times 4 = 16\); sum 5 in (1,4),(2,3),(3,2),(4,1): \(\dfrac{4}{16} = \dfrac{1}{4}\) |
| SP29 | \(\left(\frac{1}{4}\right)^5 = \dfrac{1}{1024}\) |
| SP30 | Total 6 numbers; even ones end in 2: 2 numbers, so \(\dfrac{2}{6} = \dfrac{1}{3}\) |
| SP31 | \(\dfrac{16}{40} = \dfrac{2}{5} = 0.4\) |
| SP32 | \(\dfrac{3}{60} = \dfrac{1}{20} = 0.05 = 5%\) |
Check your method
For every question: identify the experiment, write the sample space (in your head or on paper), count favourable outcomes, then simplify. A probability that is not between 0 and 1 is a sign you miscounted.
39. Chapter Checklist
- I can explain probability as a measurement and place events on the scale from 0 to 1.
- I use the words impossible, unlikely, equally likely, more likely and certain with their scale positions.
- I can compute experimental probability and relative frequency from a frequency table.
- I can compute theoretical probability and always state the equally-likely assumption first.
- I can write any probability as a fraction, decimal and percentage.
- I can list a sample space completely, without duplicates, and state \(n(S)\).
- I can identify an event as a subset \(E \subseteq S\) and use \(1 - P(E)\) for complements.
- I can build tree diagrams for two-stage experiments, multiplying along paths and adding across paths.
- I can distinguish drawing with replacement from drawing without replacement.
- I can explain why a fair coin has no memory, and identify the Gambler’s Fallacy.
- I understand the Law of Large Numbers and can use a sample to estimate a population.
- I attempted all 32 self-practice problems and checked my answers.
Well done! You have given numbers to the “maybe” world around you — measuring likelihood, listing possibilities, drawing trees and reasoning about fairness and memory. These ideas connect directly to statistics, chance and data, which you will meet again in Class 10 and beyond. Now test yourself with the practice paper, quiz and test below.