NCERT Solutions Class 9 Maths Chapter 7 Think and Reflect — The Mathematics of Maybe: Introduction to Probability

NCERT Class 9 Mathematics — Ganita Manjari Part I, The Mathematics of Maybe: Introduction to Probability. All 5 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

The five boxes in this chapter are all about the gap between knowing the list and knowing the result. A random experiment tells you everything that could happen and nothing about what will; learning to live with that gap, and to say useful things inside it, is what the chapter is for.

Key insight. Independent trials have no memory. A coin that has landed heads ten times is no more likely to give a tail next; a die that has shown four eight times running is still $\frac16$ to show it again. Probability describes the long run, and the long run has no way of reaching back to influence the next throw.

Prompt 1 comes from the box on page 156, at the end of the section on randomness.

Question 1

Such unpredictability can be useful sometimes! For example, 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?

Solution. The toss is fair because of two properties working together, and it needs both.

It is symmetric. The two outcomes are equally likely:

$$P(\text{heads}) = P(\text{tails}) = \frac12$$

so whichever side each captain calls, each has exactly the same chance of winning the toss. Neither team is favoured by the method itself.

It is unpredictable. Nobody — not the captains, not the umpires, not the groundsman — can work out in advance how the coin will land. This is what stops the method being gamed. If the result could be predicted or influenced, symmetry alone would not save it: a method where the outcome is even money but one side secretly knows it in advance is not fair at all.

There is a third, practical property worth adding: the toss is public and verifiable. Both captains and the umpires watch it, so nobody has to trust anybody’s word about the result. A fair procedure that cannot be seen to be fair does not settle an argument.

Notice what is not being claimed. The toss does not make the match fair — the team that wins it gets a real advantage on some pitches. What it makes fair is the allocation of that advantage: neither side can claim the choice was rigged, because the method has no way of preferring one over the other.

Prompt 2 comes from the box on page 157, closing the same section.

Question 2

Ask your friend to predict the outcome of a ₹1 coin you toss. Do you see that your friend could guess heads or tails but could not know for certain? That’s randomness! All possible results are known, but each individual try is unpredictable.

Solution. This box is an activity with its point already stated, so what it asks is that you carry it out and see the statement become obvious. Do it a dozen times and record the guesses against the results.

What you will find. Your friend’s guesses will be right about half the time — near $6$ out of $12$, though anything from about $3$ to $9$ is unremarkable. And crucially, no strategy improves on that. Always guessing heads, alternating heads and tails, copying the last result, avoiding the last result: all of them come out at about half, because the coin is not responding to any of them.

The two halves of randomness. The sentence in the box names both, and they pull in opposite directions:

• All possible results are known. The sample space is $S = \{H, T\}$, settled before the coin is tossed. There is no third possibility to worry about, and no uncertainty about what could happen.
• Each individual try is unpredictable. Which of the two it will be, on this particular toss, is unknowable in advance.

That combination is exactly what makes probability possible as a subject. If nothing were known, there would be nothing to compute; if everything were known, there would be nothing to compute about. Probability lives in between: it cannot tell you this toss, but it can tell you that about half of many tosses will be heads — and that is enough to decide who bats first, to price insurance, or to test a medicine.

Prompt 3 comes from the box on page 163, just after the law of large numbers is stated.

Question 3

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.

Solution. The statement is correct, and it is the single most misunderstood fact in the whole chapter. It is worth unpacking why.

Why the previous rolls do not matter. The die is an object made of plastic. It does not record what it has done, and it has no mechanism by which the ninth roll could depend on the first eight. Each roll is an independent trial, so

$$P(\text{a } 4 \text{ on the next roll}) = \frac16 \approx 0.167$$

whatever came before. (The book writes $\approx 0.16$; $\frac16$ is $0.1\overline{6}$, so $0.167$ is the closer two-figure value.)

The fallacy this kills. The instinct that a $4$ is now “less likely” — or, equally wrongly, that the die is “hot” and more likely to give another — is the gambler’s fallacy. It comes from a half-memory of the law of large numbers: people know the proportions even out eventually, and conclude that something must be pushing them back. Nothing is. The proportions even out because the early results get diluted by thousands of later ones, not because they get corrected.

A genuine subtlety. Eight fours in a row is itself extraordinarily unlikely:

$$\left(\frac16\right)^{8} = \frac{1}{1\,679\,616}$$

about one chance in $1.7$ million. So if this really happened to you, the sensible response is not to make a prediction about roll nine but to doubt the assumption — to suspect the die is loaded, or the roll is being rigged. That is why the box says “assuming the die is fair”. Given that assumption, the answer is $\frac16$; whether the assumption survives a run like that is a separate question, and a good one.

Prompt 4 comes from the box on page 167, in the section on events.

Question 4

When we used the sample space $\{\text{Rain}, \text{No Rain}\}$ in Example 1, 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 ${\text{No Rain}, \text{Drizzle}, \text{Light Rain}, \text{Heavy Rain}}$ 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.

Solution. The point being made is that a sample space is chosen, not discovered. The same tomorrow can be described by either list, and neither is wrong; they answer different questions.

Match the detail to the question.

• If you are deciding whether to carry an umbrella, $\{\text{Rain}, \text{No Rain}\}$ is enough. Any rain at all means take it.
• If you are deciding whether to hold a school sports day, drizzle and heavy rain are quite different prospects, and a two-way split throws away exactly the information you need.
• A farmer deciding when to sow would want more detail still — how many millimetres, and over how many days.

The two lists are consistent. Nothing about the finer one contradicts the coarser one; the finer one refines it:

$$\text{Rain} = \text{Drizzle} \cup \text{Light Rain} \cup \text{Heavy Rain}$$

so “Rain” survives in the four-outcome space as an event rather than as an outcome. That is the general pattern: refining a sample space turns an old outcome into a set of new ones, and every probability you had before can still be recovered by adding.

One caution the box does not state. Neither of these sample spaces has equally likely outcomes. Writing $P(\text{Heavy Rain}) = \frac14$ because the list has four entries would be badly wrong — in most places heavy rain is far rarer than no rain at all. These probabilities have to come from meteorological data, which is to say they are experimental probabilities. Listing outcomes and assuming they are equally likely are two separate steps, and only the first has been done here.

Prompt 5 comes from the box on page 169, immediately after the tree diagram for two coin tosses.

Question 5

Can you calculate the probability of getting one head and one tail?

Solution. The experiment is the one just worked in the book: a fair coin tossed twice, for which the tree diagram gives

$$S = \{HH, HT, TH, TT\}, \qquad n(S) = 4$$

and all four outcomes are equally likely, each with probability $\frac14$.

“One head and one tail” does not say which came first, so both mixed outcomes count:

$$E = \{HT, TH\}, \qquad P(E) = \frac{2}{4} = \frac{1}{2}$$

The trap. It is very natural to think of “one head and one tail” as a single result and answer $\frac14$ — the same as $HH$, which the book has just computed. It is not. $HH$ can happen in exactly one way, while a head and a tail can happen in two, and that is what doubles its probability.

Checking the whole sample space makes the structure clear:

The three probabilities add to $1$ ✓, and the middle row is twice the others precisely because two paths of the tree lead to it. This asymmetry is the whole reason a tree diagram is worth drawing: it shows the paths, and paths are what you count.

Common mistakes

• Prompt 1, saying the toss is fair “because it is random”. Randomness alone is not fairness. A biased coin is random too. Fairness needs the two outcomes to be equally likely as well as unpredictable.
• Prompt 3, believing a $4$ is now “due” — or now “hot”. Both are the gambler’s fallacy in different clothes. Independent trials carry no information forward at all.
• Prompt 3, confusing two different questions. “What is the chance of eight fours in a row?” ($\frac{1}{6^8}$, tiny) and “what is the chance of a four given eight already?” ($\frac16$) have very different answers, and mixing them up is what makes the fallacy feel convincing.
• Prompt 4, treating the four rainfall outcomes as equally likely. Four entries in a list do not make each one $\frac14$. Their probabilities come from weather records.
• Prompt 5, answering $\frac14$. $HT$ and $TH$ are two distinct outcomes, so the event has probability $\frac12$. Only $HH$ and $TT$ are single outcomes.
• Forgetting to check that the probabilities add to $1$. Every one of these answers can be sanity-checked that way in a single line, and it catches a missed outcome immediately.

Practise next

• Exercise Set 7.2, Question 3(iv) and Question 6 — the gambler’s fallacy and the law of large numbers as exercise questions, which is where prompt 3 leads directly.
• Exercise Set 7.3 — choosing and listing sample spaces, the skill prompt 4 is about.
• The End-of-Chapter Exercises, Question 3 — five experiments, and deciding for each whether its outcomes are equally likely.