NCERT Exemplar Solutions Class 10 Maths Statistics and Probability — Exercise 13.2

NCERT Exemplar Problems, Class 10 Mathematics — Statistics and Probability, Exercise 13.2. All 14 questions solved.

Fourteen questions, and twelve of them are the same mistake wearing different clothes. The mistake is this:

If a situation has $n$ possible outcomes, then each outcome has probability $\frac1n$.

That is false, and it is false because the definition of probability requires the outcomes to be equally likely:

$$P(E) = \frac{\text{number of favourable outcomes}}{\text{number of } \textbf{equally likely} \text{ outcomes}}$$

Listing outcomes is easy; making sure they are equally likely is the work. “No girl, one girl, two girls, three girls” is a perfectly good list of four outcomes, but one girl can happen three ways and no girl only one, so they are not equally likely and the probabilities are not $\frac14$ each.

The reliable method is to go back to the underlying equally likely outcomes — the individual coin sequences, the ordered pairs of dice — count them, and group them afterwards. Questions 5, 7 and 10 all fall out immediately once that list is on paper.

Three of the questions (11, 12, 13) are the gambler’s fallacy: the belief that a run of heads makes a tail “due”. A coin has no memory. Each toss is a fresh experiment with the same two equally likely outcomes, whatever happened before.

Only two answers here are “Yes” — Questions 8 and 14 — and in both the outcomes really are symmetric. That contrast is the point of the exercise.

A note on the current syllabus. Chapter 13 keeps the mean, mode and median (Questions 1 to 4 here), and Chapter 14’s treatment of probability is intact. Nothing in this exercise is affected by the rationalisation, though the ogive section that Exercise 13.1’s Question 5 depends on has gone.

Key insight. Before dividing by the number of outcomes, ask whether swapping any two of them would leave the situation looking the same. For a coin, swapping head and tail changes nothing — so they are equally likely. For “no girl” and “one girl”, no such symmetry exists, and the equal division is unjustified.

Question 1

The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

Solution. Not always, and the reason is what grouping throws away.

Once data is grouped, the individual values are lost — all you know is how many observations fell in each class. The median formula

$$\text{Median} = l + \left(\frac{\frac n2 – cf}{f}\right)h$$

recovers a value by assuming the observations inside the median class are spread evenly across it, and then interpolating to the right position.

Real data is rarely spread evenly inside a class. If the observations in the median class happen to cluster near its lower end, the true median is lower than the formula reports; cluster them near the top and it is higher.

A concrete case. Take the five values $10, 11, 12, 28, 29$. Ungrouped, the median is the middle value, $12$. Grouped into classes $10$–$20$ and $20$–$30$ with frequencies $3$ and $2$, the formula gives

$$10 + \left(\frac{2.5 – 0}{3}\right)(10) = 10 + 8.33 = 18.33$$

The two differ by more than six, because the three values in the first class sit at its very bottom rather than spread across it.

So the grouped median is an estimate, and agreement with the ungrouped value is a coincidence rather than a rule.

Question 2

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula

$$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$$

where $a$ is the assumed mean. $a$ must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

Solution. No. The assumed mean $a$ can be any number at all — the mean that comes out does not depend on it.

Here is why. By definition $d_i = x_i – a$, so

$$a + \frac{\sum f_i d_i}{\sum f_i} = a + \frac{\sum f_i (x_i – a)}{\sum f_i} = a + \frac{\sum f_i x_i – a\sum f_i}{\sum f_i} = a + \frac{\sum f_i x_i}{\sum f_i} – a$$

The two $a$’s cancel, leaving

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

which is the direct-method mean and contains no $a$ whatsoever. Whatever value of $a$ you choose, the same $\bar{x}$ emerges.

So why choose a class mark? Purely for arithmetic convenience. Picking $a$ to be a central class mark makes the deviations $d_i$ small, symmetric about zero, and often divisible by the class width — which is what makes the step-deviation method of Exercise 13.1’s Question 4 worth using. Choosing $a = 1000$ for data around $50$ would work perfectly and produce ugly numbers.

Convenience is not a requirement. The statement confuses “usually chosen this way” with “must be”.

Question 3

Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

Solution. No. The three can coincide, and for one important family of distributions they always do.

When a distribution is symmetric and single-peaked, the mean, median and mode all fall at the centre:

• the mode is at the peak, which is the centre by symmetry
• the median splits the data in half, and symmetry puts that split at the centre
• the mean balances the deviations, and symmetry balances them at the centre

A small example. Take classes $0$–$10$, $10$–$20$, $20$–$30$ with frequencies $5$, $10$, $5$. The mode is in $10$–$20$; $n = 20$ so $\frac n2 = 10$ puts the median there too; and the mean is $\frac{5(5) + 10(15) + 5(25)}{20} = \frac{300}{20} = 15$, the mid-point of that same class.

All three land on $15$.

They differ when the data is skewed — a few very large values drag the mean up while leaving the median and mode behind, which is why average incomes exceed median incomes almost everywhere. The relation $\text{Mode} = 3\,\text{Median} – 2\,\text{Mean}$ is the empirical rule describing that gap, and it correctly gives all three equal when any two are.

Question 4

Will the median class and modal class of grouped data always be different? Justify your answer.

Solution. Not always — they can be the same class, and it is easy to see when.

The median class is the first whose cumulative frequency reaches $\frac n2$. The modal class is the one with the largest frequency. These are different criteria, but nothing prevents one class from satisfying both.

A large frequency both makes a class the modal class and pushes the cumulative total past $\frac n2$ within it — so for data that peaks near the middle, the two usually coincide.

A worked instance is already in this unit. Question 9 of Exercise 13.1 has classes $65$–$85$ through $185$–$205$ with frequencies $4, 5, 13, 20, 14, 7, 4$. There:

• $n = 67$, so $\frac n2 = 33.5$; cumulative frequencies are $4, 9, 22, 42, \ldots$, so the median class is $125$–$145$
• the largest frequency is $20$, also in $125$–$145$

Same class. And that is why the question there asks for the difference between its upper and lower limits rather than between two different classes.

They differ, by contrast, in Question 6 of the same exercise — median class $10$–$15$, modal class $15$–$20$.

Question 5

In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $\frac14$. Is this correct? Justify your answer.

Solution. No. The four outcomes listed are real, but they are not equally likely, so they cannot each have probability $\frac14$.

Go back to the underlying outcomes. Each child is a boy or a girl, and the three children give $2^3 = 8$ equally likely sequences:

$$BBB, \quad BBG, \ BGB, \ GBB, \quad BGG, \ GBG, \ GGB, \quad GGG$$

Now group them by the number of girls:

Number of girls	Sequences	Count	Probability
$0$	$BBB$	1	$\frac18$
$1$	$BBG$, $BGB$, $GBB$	3	$\frac38$
$2$	$BGG$, $GBG$, $GGB$	3	$\frac38$
$3$	$GGG$	1	$\frac18$

So the probabilities are $\frac18, \frac38, \frac38, \frac18$ — not $\frac14$ each. They do sum to $1$, as they must.

Why “one girl” is three times as likely as “no girl” is the heart of it: there is only one way to have no girls, but three different children could each be the one girl. Grouping outcomes together adds their probabilities, and groups of different sizes get different totals.

Question 6

A game consists of spinning an arrow which comes to rest pointing at one of the regions ($1$, $2$ or $3$) (Fig. 13.1). Are the outcomes $1$, $2$ and $3$ equally likely to occur? Give reasons.

Solution. No. Read the figure carefully: the circle is cut by a full horizontal diameter and by a single vertical radius running only from the top down to the centre.

So the three regions are:

• region 1 — the upper-left quadrant, a quarter of the circle
• region 2 — the upper-right quadrant, a quarter
• region 3 — the entire lower semicircle, a half

For a spinner, the chance of stopping in a region is proportional to that region’s share of the circle, so

$$P(1) = \frac14, \qquad P(2) = \frac14, \qquad P(3) = \frac12$$

These sum to $1$ ✓, but they are not equal. The outcome $3$ is twice as likely as either of the others.

The lesson is that “three outcomes” says nothing about their likelihood. Here the count of regions is three and the areas are in the ratio $1 : 1 : 2$. Had the vertical line been a full diameter, there would have been four equal quadrants and a genuinely fair spinner.

Question 7

Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number $36$? Why?

Solution. Work out each probability separately.

Apoorv. Two dice give $6 \times 6 = 36$ equally likely ordered pairs. Which have product $36$? The largest product available is $6 \times 6 = 36$, and that is reached only one way:

$$(6, 6)$$

No other pair works — $36 = 4 \times 9$ or $3 \times 12$ would need a face above $6$. So

$$P(\text{Apoorv}) = \frac{1}{36}$$

Peehu. One die gives $6$ equally likely outcomes. Her result is $36$ when the die shows $6$, since $6^2 = 36$ — and only then.

$$P(\text{Peehu}) = \frac16$$

Compare on a common denominator:

$$P(\text{Apoorv}) = \frac{1}{36}, \qquad P(\text{Peehu}) = \frac16 = \frac{6}{36}$$

Peehu’s chance is six times Apoorv’s.

The reason is that Apoorv needs two sixes and Peehu needs only one. Each six has probability $\frac16$, and requiring a second one multiplies by $\frac16$ again.

Question 8

When we toss a coin, there are two possible outcomes — Head or Tail. Therefore, the probability of each outcome is $\frac12$. Justify your answer.

Solution. Yes, this one is correct — and it is worth being clear about why, because the same reasoning fails in Questions 5, 9, 10 and 14’s neighbours.

The two outcomes are equally likely, and that is a claim about the coin, not about the number of outcomes. A fair coin is physically symmetric: the two faces are alike in mass and shape, so nothing distinguishes head from tail. Swapping the labels would leave the situation identical.

Given equal likelihood, the two probabilities are equal and sum to $1$, so each is $\frac12$.

The contrast with Question 9. There the two outcomes are “$1$” and “not $1$” on a die, which are also two in number — but they are not symmetric, because “not $1$” gathers five faces and “$1$” only one. No relabelling leaves that situation unchanged.

So the count of outcomes is never the reason; the symmetry is. Two outcomes give $\frac12$ each only when the two are interchangeable, and a coin is the standard case where they are.

Question 9

A student says that if you throw a die, it will show up $1$ or “not $1$”. Therefore, the probability of getting $1$ and the probability of getting “not $1$” each is equal to $\frac12$. Is this correct? Give reasons.

Solution. No. Two outcomes, but nothing like equally likely.

A die has six equally likely faces. Of these:

• “$1$” is favoured by exactly one face: $\{1\}$
• “not $1$” is favoured by five: $\{2, 3, 4, 5, 6\}$

$$P(1) = \frac16, \qquad P(\text{not } 1) = \frac56$$

They sum to $1$, as a pair of complementary events must, but they are nowhere near equal — “not $1$” is five times as likely.

Read this directly against Question 8. Both describe a situation with two outcomes; one gives $\frac12$ each and the other does not. The difference is not the count but the symmetry: head and tail are interchangeable, whereas “$1$” and “not $1$” are built from one face and five.

The student’s error is treating “the situation has two outcomes” as though it implied “each has probability $\frac12$”. Any event and its complement give two outcomes — “it rains tomorrow” or “it does not” — and almost never with equal probability.

Question 10

I toss three coins together. The possible outcomes are no heads, $1$ head, $2$ heads and $3$ heads. So, I say that probability of no heads is $\frac14$. What is wrong with this conclusion?

Solution. The list of four outcomes is correct; treating them as equally likely is not.

Three coins give $2^3 = 8$ equally likely sequences:

$$TTT, \quad HTT, \ THT, \ TTH, \quad HHT, \ HTH, \ THH, \quad HHH$$

Grouping by the number of heads:

Heads	Sequences	Probability
$0$	$TTT$	$\frac18$
$1$	$HTT$, $THT$, $TTH$	$\frac38$
$2$	$HHT$, $HTH$, $THH$	$\frac38$
$3$	$HHH$	$\frac18$

So

$$P(\text{no heads}) = \frac18, \quad \text{not } \frac14$$

What is wrong, stated precisely: the four described outcomes are not equally likely, so dividing $1$ among them equally is unjustified. “No heads” can happen in only one way, while “$1$ head” can happen in three.

This is Question 5 with coins in place of children — the same $8$-outcome sample space and the same $1 : 3 : 3 : 1$ pattern. Recognising that pattern is what makes both questions immediate.

Question 11

If you toss a coin $6$ times and it comes down heads on each occasion. Can you say that the probability of getting a head is $1$? Give reasons.

Solution. No. The probability of a head remains $\frac12$, whatever the last six tosses did.

A coin has no memory. Each toss is an independent experiment, and the coin’s physical symmetry — the thing that makes head and tail equally likely — is unaffected by what happened before. Nothing about the coin changed during those six tosses.

Six heads in a row is unremarkable. Its probability is

$$\left(\frac12\right)^6 = \frac{1}{64}$$

which is about $1.6\%$ — uncommon, but the sort of thing that happens regularly if you toss coins often. In $64$ attempts at six tosses you would expect it about once.

What a probability of $1$ would mean. It would mean a head is certain — that tails is impossible. Six observations cannot establish that, and the very next toss could refute it.

The confusion is between the theoretical probability, fixed by the coin’s symmetry at $\frac12$, and the experimental frequency, which is $\frac66 = 1$ in this short run and settles towards $\frac12$ as the number of tosses grows.

Question 12

Sushma tosses a coin $3$ times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

Solution. It could be either, and each is equally likely. The next toss is a head with probability $\frac12$ and a tail with probability $\frac12$, exactly as the first one was.

The three tails already observed have no influence. The coin is unchanged, and each toss is independent of the ones before it.

Two opposite wrong instincts, and this question and the next are built to catch one each:

• “It is on a run of tails, so a tail is more likely.” No — the coin cannot know it is on a run.
• “Three tails already, so a head is due.” Also no, and that is Question 13.

Neither is right, because the coin’s behaviour on the fourth toss is determined entirely by its symmetry and not at all by its history. Believing otherwise is the gambler’s fallacy, and it is the most persistent misconception in elementary probability.

Question 13

If I toss a coin $3$ times and get head each time, should I expect a tail to have a higher chance in the $4$th toss? Give reason in support of your answer.

Solution. No. Head and tail are equally likely on the fourth toss, so a tail has probability $\frac12$ — no higher than it ever was.

This is the gambler’s fallacy in its commonest form: the belief that a run of one outcome makes the other “due”, as though the coin were keeping a tally and correcting for it.

Why the intuition is tempting, and where it goes wrong. Over many tosses the proportion of heads does settle towards $\frac12$ — that much is true. But the mechanism is not that later tosses compensate for earlier ones. It is that the early results are swamped by the sheer number of later ones.

After $3$ heads the running proportion is $1$. After a further $997$ tosses splitting roughly evenly, it is about $\frac{3 + 499}{1000} \approx 0.502$ — close to $\frac12$ not because any toss was biased towards tails, but because three extra heads among a thousand tosses hardly registers.

The imbalance is diluted, never repaid.

Question 14

A bag contains slips numbered from $1$ to $100$. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is $\frac12$. Justify.

Solution. Yes — the conclusion is correct here. But the reasoning given is not, and the question asks you to justify, which means supplying the right reason.

The reason offered is wrong. “The situation has only two possible outcomes, so each has probability $\frac12$” is exactly the fallacy Questions 5, 9 and 10 expose. Question 9’s die also has two outcomes, “$1$” and “not $1$”, with probabilities $\frac16$ and $\frac56$.

The correct justification is the count. Among the numbers $1$ to $100$:

• odd numbers: $1, 3, 5, \ldots, 99$ — there are $50$
• even numbers: $2, 4, 6, \ldots, 100$ — there are $50$

$$P(\text{odd}) = \frac{50}{100} = \frac12, \qquad P(\text{even}) = \frac{50}{100} = \frac12$$

So the two are equally likely because the two counts are equal, not because there are two of them.

The distinction has consequences. Had the bag held slips $1$ to $99$, there would be $50$ odd and $49$ even, giving $\frac{50}{99}$ and $\frac{49}{99}$ — still two outcomes, no longer equal. The right answer with the wrong reason survives only as long as the numbers happen to cooperate.

Common mistakes

• Questions 5, 9, 10 and 14 — assuming $n$ outcomes means $\frac1n$ each. The definition of probability requires the outcomes to be equally likely. Go back to the underlying symmetric outcomes — the individual coin sequences, the six faces, the hundred slips — count those, and group afterwards.
• Questions 11, 12 and 13 — the gambler’s fallacy. A coin has no memory, so previous results never change the next toss. Long-run proportions settle by dilution, not by compensation.
• Question 14 — giving the right answer with the wrong reason. The probabilities are $\frac12$ because $50 = 50$, not because there are two outcomes. With slips $1$ to $99$ the same reasoning would give the wrong answer.
• Question 6 — counting regions instead of measuring them. On a spinner the probability is proportional to the region’s share of the circle. Three regions here occupy $\frac14$, $\frac14$ and $\frac12$.
• Question 2 — believing the assumed mean must be a class mark. The $a$ cancels algebraically, so any value gives the same mean. Class marks are chosen for convenience alone.
• Questions 3 and 4 — reading “usually different” as “always different”. Mean, median and mode coincide for symmetric data, and the median class and modal class coincide whenever the peak sits where the cumulative frequency crosses $\frac n2$.
• Question 1 — expecting the grouped median to be exact. It is an estimate built on the assumption of uniform spread within the median class, and it can be several units out when the data clusters at one end of that class.
• Question 7 — comparing probabilities with different denominators. $\frac1{36}$ against $\frac16$ is much clearer as $\frac1{36}$ against $\frac6{36}$. Put fractions on a common denominator before declaring a winner.

Practise next

• Exercise 13.3 — forty-two short-answer questions, twenty on grouped data and twenty-two on probability, where these principles get applied to real calculations.
• Exercise 13.1 — worth revisiting, particularly Questions 12 to 17, which state as multiple-choice the range and complement facts this exercise reasons about.
• Exercise 11.2 — a different subject, but the same discipline of justifying a verdict rather than merely stating it.