NCERT Exemplar Solutions Class 10 Maths Statistics and Probability — Exercise 13.3 Q19–42

NCERT Exemplar Problems, Class 10 Mathematics — Statistics and Probability, Exercise 13.3, Questions 19 to 42. All the probability questions solved.

The second half of Exercise 13.3 — twenty-four questions on dice, cards, coins and counting. The grouped-data half, Questions 1 to 18, is on the first part.

Why there is no Exercise 13.4 page. This unit also has a Long Answer exercise, and it is not solved on this site — not by oversight. NCERT’s published PDF of the chapter stops at printed page $174$, part-way through the first worked example of the Long Answer section, and printed pages $175$ to $182$ — which carry the rest of that section and the whole of Exercise 13.4 — are not published anywhere in the Class 10 Maths exemplar directory. The file itself is intact; the chapter is not. We re-downloaded it (byte-identical), tried every plausible sibling code (all 404) and checked Wayback snapshots from 2020, 2024 and 2026 (all the same 22-page file). The answer key does print answers for its Questions 1, 2, 3, 4 and 7, so the exercise exists in the printed book — only its pages never went online. Since every question here is read from the printed page rather than reconstructed backwards from an answer, those questions cannot be written up honestly, and are left out.

Everything runs on one formula,

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

so the whole difficulty is counting correctly. Three sample spaces recur and are worth knowing cold:

• two dice — $6 \times 6 = 36$ ordered pairs. $(2,5)$ and $(5,2)$ are different outcomes, and forgetting that halves most answers.
• three coins — $2^3 = 8$ sequences: $HHH$, $HHT$, $HTH$, $THH$, $HTT$, $THT$, $TTH$, $TTT$.
• a deck — $52$ cards, $4$ suits of $13$; $26$ red and $26$ black; $12$ face cards (J, Q, K in each suit).

Two habits save most of the marks. Use the complement when the unfavourable outcomes are easier to count — Questions 19(ii), 22, 27(iii) and 31(ii) are all quicker that way. And when a question removes cards or conditions on an earlier event, recompute the total: Questions 28, 29, 30, 36 and 42 all change the denominator.

A note on the current syllabus. Chapter 14 is intact — Probability: A Theoretical Approach, and a Summary. Nothing in these twenty-four questions is affected by the rationalisation.

Key insight. Write the sample space down before counting anything. Two dice are $36$ ordered pairs, not $21$ unordered ones; three coins are $8$ sequences, not $4$ counts of heads. Almost every wrong answer in this exercise comes from collapsing a sample space that should have stayed expanded.

Question 19

Two dice are thrown at the same time. Find the probability of getting (i) same number on both dice, (ii) different numbers on both dice.

(Exercise 13.3, Question 19.)

Solution. The sample space is the $36$ ordered pairs $(a, b)$ with $a, b$ from $1$ to $6$.

(i) Same number on both. The favourable outcomes are the six “doubles”:

$$(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$$

$$P = \frac{6}{36} = \frac16$$

(ii) Different numbers. This is the complement of (i), so no separate count is needed:

$$P = 1 – \frac16 = \frac56$$

Counting directly would work too — $36 – 6 = 30$ favourable, and $\frac{30}{36} = \frac56$ ✓ — but the complement is faster and less error-prone.

Why $36$ and not $21$. If the dice were indistinguishable you might list only the $21$ unordered pairs. But $(2,5)$ and $(5,2)$ are physically different outcomes and equally likely, whereas the “pair $\{2,5\}$” is twice as likely as the “pair $\{3,3\}$”. Keeping the pairs ordered keeps them equally likely, which is what the formula requires.

Question 20

Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is (i) $7$? (ii) a prime number? (iii) $1$?

(Exercise 13.3, Question 20.)

Solution. Again $36$ ordered pairs.

(i) Sum $7$. The pairs are

$$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$$

— six of them, one for each value of the first die.

$$P = \frac{6}{36} = \frac16$$

$7$ is the commonest total precisely because it can be made in more ways than any other.

(ii) A prime sum. Possible sums run from $2$ to $12$; the primes among them are $2, 3, 5, 7, 11$. Count the ways for each:

Sum	Ways	Count
2	$(1,1)$	1
3	$(1,2), (2,1)$	2
5	$(1,4), (2,3), (3,2), (4,1)$	4
7	six pairs, as above	6
11	$(5,6), (6,5)$	2
Total		15

$$P = \frac{15}{36} = \frac{5}{12}$$

Note that $9$ is not prime ($9 = 3 \times 3$), and including its four pairs would give $\frac{19}{36}$.

(iii) Sum $1$. The smallest possible total is $1 + 1 = 2$, so a sum of $1$ is impossible:

$$P = 0$$

Question 21

Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is (i) $6$ (ii) $12$ (iii) $7$.

(Exercise 13.3, Question 21.)

Solution.

(i) Product $6$. List the ordered pairs whose product is $6$:

$$(1,6), (6,1), (2,3), (3,2)$$

$$P = \frac{4}{36} = \frac19$$

(ii) Product $12$.

$$(2,6), (6,2), (3,4), (4,3)$$

$$P = \frac{4}{36} = \frac19$$

$(12, 1)$ is not available — no die shows $12$ — and $(1,12)$ likewise.

(iii) Product $7$. Since $7$ is prime, the only factor pairs are $1 \times 7$ and $7 \times 1$, and neither works because no die shows $7$. So the event is impossible:

$$P = 0$$

The lesson from (iii), and it is the reason the question includes it: a product can only be made from factors that actually appear on a die. Every prime above $6$ — that is, $7$ and $11$ — is unreachable as a product, though both are perfectly reachable as a sum, as Question 20 showed.

Question 22

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than $9$.

(Exercise 13.3, Question 22.)

Solution. Go through the first die one value at a time, and ask which second values keep the product under $9$.

First die	Second die values with product $< 9$	Count
1	$1, 2, 3, 4, 5, 6$	6
2	$1, 2, 3, 4$	4
3	$1, 2$	2
4	$1, 2$	2
5	$1$	1
6	$1$	1
Total		16

$$P = \frac{16}{36} = \frac49$$

Two places to be careful. With the first die $2$, the product $2 \times 4 = 8$ is less than $9$ so it counts, but $2 \times 5 = 10$ does not. And with the first die $3$, $3 \times 3 = 9$ is not less than $9$ — “less than” excludes equality, and that single pair is the difference between $\frac{16}{36}$ and $\frac{17}{36}$.

Working row by row like this is more reliable than trying to spot the pattern, because the cut-off falls in a different place in each row.

Question 23

Two dice are numbered $1, 2, 3, 4, 5, 6$ and $1, 1, 2, 2, 3, 3$, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from $2$ to $9$ separately.

(Exercise 13.3, Question 23.)

Solution. The second die is not standard — it shows $1, 1, 2, 2, 3, 3$, so each of $1$, $2$, $3$ appears twice.

There are still $36$ equally likely outcomes, because each die still has six faces. The sums run from $1 + 1 = 2$ to $6 + 3 = 9$.

For a given sum $s$, count the pairs: the first die must show $a$ and the second $s – a$, and the second die offers two faces for each of $1$, $2$, $3$.

Sum	Possible (first, second)	Ways	Probability
2	$(1,1)$	$1 \times 2 = 2$	$\frac{2}{36} = \frac{1}{18}$
3	$(1,2), (2,1)$	$2 + 2 = 4$	$\frac{4}{36} = \frac19$
4	$(1,3), (2,2), (3,1)$	$2 + 2 + 2 = 6$	$\frac{6}{36} = \frac16$
5	$(2,3), (3,2), (4,1)$	$6$	$\frac16$
6	$(3,3), (4,2), (5,1)$	$6$	$\frac16$
7	$(4,3), (5,2), (6,1)$	$6$	$\frac16$
8	$(5,3), (6,2)$	$4$	$\frac19$
9	$(6,3)$	$2$	$\frac{1}{18}$

Check: the eight probabilities sum to

$$\frac{1}{18} + \frac19 + \frac16 \times 4 + \frac19 + \frac{1}{18} = \frac{2 + 4 + 24 + 4 + 2}{36} = 1 \ \checkmark$$

The distribution is symmetric about $5.5$ and flat across the middle four sums — a shape quite unlike the triangular one two standard dice give. Repeating a face on one die changes the whole profile.

Question 24

A coin is tossed two times. Find the probability of getting at most one head.

(Exercise 13.3, Question 24.)

Solution. Two tosses give $2^2 = 4$ equally likely sequences:

$$HH, \quad HT, \quad TH, \quad TT$$

“At most one head” means zero heads or one head — so everything except $HH$:

$$TT, \quad HT, \quad TH$$

$$P = \frac34$$

By the complement, more quickly: the only outcome with more than one head is $HH$, so

$$P = 1 – \frac14 = \frac34$$

Reading the phrase correctly is the whole question:

• “at most one head” — $0$ or $1$ head — $\frac34$
• “at least one head” — $1$ or $2$ — $\frac34$ as it happens
• “exactly one head” — $HT$ or $TH$ — $\frac24 = \frac12$

The first two coincide here only by coincidence of two tosses; with three coins they differ.

Question 25

A coin is tossed $3$ times. List the possible outcomes. Find the probability of getting (i) all heads (ii) at least $2$ heads.

(Exercise 13.3, Question 25.)

Solution. The eight possible outcomes, which the question asks for explicitly:

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

These are equally likely; the numbers of heads are not, as Question 10 of Exercise 13.2 established.

(i) All heads. One outcome, $HHH$:

$$P = \frac18$$

(ii) At least $2$ heads means $2$ or $3$ heads:

• two heads: $HHT$, $HTH$, $THH$ — three outcomes
• three heads: $HHH$ — one outcome

$$P = \frac{3 + 1}{8} = \frac48 = \frac12$$

A symmetry check on (ii). “At least $2$ heads” and “at least $2$ tails” are mirror images, so they must have equal probability; and between them they cover every outcome exactly once, since with three coins you cannot have two of each. So each is $\frac12$ ✓ — which confirms the count without recounting.

Question 26

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is $2$.

(Exercise 13.3, Question 26.)

Solution. “Difference is $2$” means the two numbers are two apart, in either order — so $(1,3)$ and $(3,1)$ both count.

List them by the smaller number:

$$(1,3), (3,1), \quad (2,4), (4,2), \quad (3,5), (5,3), \quad (4,6), (6,4)$$

That is four unordered pairs, each appearing twice:

$$4 \times 2 = 8$$

$$P = \frac{8}{36} = \frac29$$

Counting only one order — listing $(1,3), (2,4), (3,5), (4,6)$ and stopping — gives $\frac{4}{36} = \frac19$, exactly half the right answer. The sample space is ordered pairs, so both orders must be counted.

The upper limit is worth noting too: the difference cannot exceed $5$, so there are only four starting values, not six. $(5,7)$ and $(6,8)$ do not exist.

Question 27

A bag contains $10$ red, $5$ blue and $7$ green balls. A ball is drawn at random. Find the probability of this ball being a (i) red ball (ii) green ball (iii) not a blue ball.

(Exercise 13.3, Question 27.)

Solution. The total number of balls is

$$10 + 5 + 7 = 22$$

(i) Red:

$$P = \frac{10}{22} = \frac{5}{11}$$

(ii) Green:

$$P = \frac{7}{22}$$

(iii) Not blue. By the complement, which is quicker than adding red and green:

$$P = 1 – \frac{5}{22} = \frac{17}{22}$$

Directly: $10 + 7 = 17$ balls are not blue, giving the same $\frac{17}{22}$ ✓.

A check that catches an arithmetic slip. The three colours are exhaustive and mutually exclusive, so

$$P(\text{red}) + P(\text{blue}) + P(\text{green}) = \frac{10 + 5 + 7}{22} = 1 \ \checkmark$$

Always confirm the total before computing anything: reading “10 red, 5 blue and 7 green” as $21$ balls would make every answer wrong by a little, and nothing in the arithmetic would object.

Question 28

The king, queen and jack of clubs are removed from a deck of $52$ playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is (i) a heart (ii) a king.

(Exercise 13.3, Question 28.)

Solution. Recompute the total first. Three cards are removed, so

$$52 – 3 = 49 \text{ cards remain}$$

(i) A heart. The removed cards are all clubs, so the hearts are untouched — all $13$ remain:

$$P = \frac{13}{49}$$

(ii) A king. A deck has four kings, one per suit, and the king of clubs has been removed. Three remain:

$$P = \frac{3}{49}$$

Both parts turn on which suit was removed. The hearts are unaffected because nothing removed was a heart; the kings lose exactly one because exactly one king was among the removed cards.

The commonest error is leaving the denominator at $52$, which would give $\frac{13}{52} = \frac14$ and $\frac{3}{52}$ — both wrong, and the first suspiciously tidy. Whenever cards are removed, written on, or set aside, the denominator changes with them.

Question 29

Refer to Q.28. What is the probability that the card is (i) a club (ii) $10$ of hearts?

(Exercise 13.3, Question 29.)

Solution. The same reduced deck: $49$ cards, with the king, queen and jack of clubs removed.

(i) A club. A full suit has $13$ cards and three clubs have gone:

$$13 – 3 = 10 \text{ clubs remain}$$

$$P = \frac{10}{49}$$

(ii) The $10$ of hearts. That is one specific card, and it was not among those removed, so it is still there:

$$P = \frac{1}{49}$$

Part (ii) is a reading test. “$10$ of hearts” names a single card, not the ten hearts and not the four tens. Compare Question 21 of Exercise 13.1, which turned on exactly the same phrase — X of Y in a card question always picks out one card.

Question 30

All the jacks, queens and kings are removed from a deck of $52$ playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value $1$ similar value for other cards, find the probability that the card has a value (i) $7$ (ii) greater than $7$ (iii) less than $7$.

(Exercise 13.3, Question 30.)

Solution. Count what is removed. Jacks, queens and kings — three ranks, four suits each:

$$3 \times 4 = 12 \text{ cards removed}, \qquad 52 – 12 = 40 \text{ remain}$$

The remaining cards have values $1$ (ace) through $10$, four of each. That is a tidy sample space: $10$ values $\times$ $4$ suits $= 40$ ✓.

(i) Value $7$. Four sevens:

$$P = \frac{4}{40} = \frac{1}{10}$$

(ii) Greater than $7$. Values $8$, $9$, $10$ — three ranks:

$$P = \frac{3 \times 4}{40} = \frac{12}{40} = \frac{3}{10}$$

(iii) Less than $7$. Values $1$ through $6$ — six ranks:

$$P = \frac{6 \times 4}{40} = \frac{24}{40} = \frac35$$

The three add to $1$:

$$\frac{1}{10} + \frac{3}{10} + \frac35 = \frac{1 + 3 + 6}{10} = 1 \ \checkmark$$

They must, because every remaining card has a value that is less than $7$, equal to $7$, or greater than $7$. That check would catch the commonest slip here — counting the ace as $11$ or $14$, or forgetting it counts at all, either of which breaks the total.

Question 31

An integer is chosen between $0$ and $100$. What is the probability that it is (i) divisible by $7$? (ii) not divisible by $7$?

(Exercise 13.3, Question 31.)

Solution. “Between $0$ and $100$” excludes both endpoints, so the integers available are

$$1, 2, 3, \ldots, 99$$

That is $99$ integers — not $100$ and not $101$.

(i) Divisible by $7$. The multiples of $7$ in that range are

$$7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98$$

There are $\left\lfloor \frac{99}{7} \right\rfloor = 14$ of them.

$$P = \frac{14}{99}$$

(ii) Not divisible by $7$. By the complement:

$$P = 1 – \frac{14}{99} = \frac{85}{99}$$

The denominator is the whole question. Reading “between $0$ and $100$” as inclusive would give $101$ integers and bring in $0$, which is divisible by $7$ — producing $\frac{15}{101}$. The key’s $\frac{14}{99}$ confirms the exclusive reading, and that is the convention “between” carries in these questions.

Question 32

Cards with numbers $2$ to $101$ are placed in a box. A card is selected at random. Find the probability that the card has (i) an even number (ii) a square number.

(Exercise 13.3, Question 32.)

Solution. From $2$ to $101$ inclusive there are

$$101 – 2 + 1 = 100 \text{ cards}$$

(i) An even number. The evens are $2, 4, 6, \ldots, 100$ — that is $50$ of them:

$$P = \frac{50}{100} = \frac12$$

The range starts at an even number and ends at an odd one, so the two parities split exactly.

(ii) A square number. List the perfect squares in $[2, 101]$:

$$4, 9, 16, 25, 36, 49, 64, 81, 100$$

— that is $2^2$ through $10^2$, nine of them.

$$P = \frac{9}{100}$$

$1$ is not in the range, which is why there are nine and not ten. The cards start at $2$, so $1^2 = 1$ is excluded; and $11^2 = 121$ is past $101$. Both ends have to be checked.

Question 33

A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.

(Exercise 13.3, Question 33.)

Solution. The English alphabet has $26$ letters, of which five are vowels:

$$A, \ E, \ I, \ O, \ U$$

The rest are consonants:

$$26 – 5 = 21$$

$$P = \frac{21}{26}$$

The letter $Y$ is counted as a consonant here, which is the convention this question assumes. In English it sometimes behaves as a vowel (“rhythm”, “my”), and had it been counted as one the answer would be $\frac{20}{26} = \frac{10}{13}$. The printed answer of $\frac{21}{26}$ settles which convention is intended.

That is worth naming in an answer rather than passing over — a question about letters is only well-posed once the vowel list is fixed.

Question 34

There are $1000$ sealed envelopes in a box, $10$ of them contain a cash prize of ₹$100$ each, $100$ of them contain a cash prize of ₹$50$ each and $200$ of them contain a cash prize of ₹$10$ each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?

(Exercise 13.3, Question 34.)

Solution. Count the envelopes that do contain a prize, then subtract.

$$10 + 100 + 200 = 310 \text{ envelopes with a prize}$$

$$1000 – 310 = 690 \text{ with no prize}$$

$$P = \frac{690}{1000} = 0.69$$

The prize amounts are irrelevant, and that is the point of listing three of them. Whether an envelope holds ₹$100$ or ₹$10$ makes no difference to whether it holds a prize; only the counts matter. Weighting the envelopes by their value would answer a different question — the expected winnings — which is not asked and is not Class 10 material.

So $69\%$ of the envelopes are empty, and the answer is most naturally given as the decimal the key uses.

Question 35

Box A contains $25$ slips of which $19$ are marked ₹$1$ and other are marked ₹$5$ each. Box B contains $50$ slips of which $45$ are marked ₹$1$ each and others are marked ₹$13$ each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than ₹$1$?

(Exercise 13.3, Question 35.)

Solution. Work out each box’s contents, then pool them.

Box A: $25$ slips, $19$ marked ₹$1$, so

$$25 – 19 = 6 \text{ marked } ₹5$$

Box B: $50$ slips, $45$ marked ₹$1$, so

$$50 – 45 = 5 \text{ marked } ₹13$$

The third box holds everything:

$$25 + 50 = 75 \text{ slips}$$

Slips marked other than ₹$1$ are the ₹$5$s and the ₹$13$s together:

$$6 + 5 = 11$$

$$P = \frac{11}{75}$$

Check by complement: slips marked ₹$1$ number $19 + 45 = 64$, and $\frac{64}{75} + \frac{11}{75} = 1$ ✓.

The amounts ₹$5$ and ₹$13$ are again irrelevant — only “is it ₹$1$ or not” matters. And the two boxes must be pooled before computing, not averaged afterwards: $\frac{6}{25}$ and $\frac{5}{50}$ do not combine by averaging, because the boxes contribute unequal numbers of slips.

Question 36

A carton of $24$ bulbs contain $6$ defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

(Exercise 13.3, Question 36.)

Solution. Two parts, and the second changes both the numerator and the denominator.

First draw — not defective. Of the $24$ bulbs, $6$ are defective and

$$24 – 6 = 18$$

are not.

$$P(\text{not defective}) = \frac{18}{24} = \frac34$$

Second draw — defective, given the first was defective and not replaced. Update the carton:

• one bulb has gone, so $24 – 1 = 23$ remain
• it was a defective one, so $6 – 1 = 5$ defective remain

$$P(\text{second defective}) = \frac{5}{23}$$

Both numbers change, and forgetting either is the usual error. Reducing only the total gives $\frac{6}{23}$; reducing only the defectives gives $\frac{5}{24}$. The bulb removed was defective, so it must be subtracted from both counts.

This is the same “without replacement” structure as Question 42’s second part, and the discipline is identical: rewrite the contents of the box before computing the next probability.

Question 37

A child’s game has $8$ triangles of which $3$ are blue and rest are red, and $10$ squares of which $6$ are blue and rest are red. One piece is lost at random. Find the probability that it is a (i) triangle (ii) square (iii) square of blue colour (iv) triangle of red colour.

(Exercise 13.3, Question 37.)

Solution. Tabulate the pieces before answering anything:

	Blue	Red	Total
Triangles	3	$8 – 3 = 5$	8
Squares	6	$10 – 6 = 4$	10
Total	9	9	18

(i) A triangle:

$$P = \frac{8}{18} = \frac49$$

(ii) A square:

$$P = \frac{10}{18} = \frac59$$

(iii) A blue square:

$$P = \frac{6}{18} = \frac13$$

(iv) A red triangle:

$$P = \frac{5}{18}$$

The table is what makes this quick. Parts (iii) and (iv) each need a single cell, and the “red” counts have to be derived — $5$ red triangles and $4$ red squares are never stated, only implied by “rest are red”.

Two checks. Parts (i) and (ii) are complementary and sum to $1$ ✓. And the four cells $3 + 5 + 6 + 4 = 18$ ✓, matching the total number of pieces.

Incidentally the colours split evenly, nine blue and nine red, though the question never asks for that.

Question 38

In a game, the entry fee is ₹$5$. The game consists of a tossing a coin $3$ times. If one or two heads show, Sweta gets her entry fee back. If she throws $3$ heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she (i) loses the entry fee (ii) gets double entry fee (iii) just gets her entry fee.

(Exercise 13.3, Question 38.)

Solution. Three coins give $8$ equally likely sequences. Sort them by the rule.

Heads	Sequences	Count	Outcome for Sweta
0	$TTT$	1	loses the fee
1	$HTT$, $THT$, $TTH$	3	gets fee back
2	$HHT$, $HTH$, $THH$	3	gets fee back
3	$HHH$	1	double the fee

(i) Loses the entry fee — that is “otherwise”, meaning neither one or two heads nor three heads. The only case left is no heads:

$$P = \frac18$$

(ii) Gets double entry fee — three heads:

$$P = \frac18$$

(iii) Just gets her entry fee — one or two heads:

$$P = \frac{3 + 3}{8} = \frac68 = \frac34$$

Check: $\frac18 + \frac18 + \frac34 = 1$ ✓ — the three outcomes are exhaustive, as they must be.

“Otherwise” has to be worked out, not read. The rules name one or two heads and three heads explicitly; everything else loses, and “everything else” turns out to be the single sequence $TTT$. Assuming “otherwise” covers several cases is the trap.

The game is fair to Sweta three-quarters of the time, and she gains only on the $\frac18$ chance of three heads.

Question 39

A die has its six faces marked $0$, $1$, $1$, $1$, $6$, $6$. Two such dice are thrown together and the total score is recorded. (i) How many different scores are possible? (ii) What is the probability of getting a total of $7$?

(Exercise 13.3, Question 39.)

Solution. Each die shows only three distinct values — $0$, $1$ and $6$ — but with multiplicities $1$, $3$ and $2$. There are still $36$ equally likely outcomes.

(i) How many different scores. Take every pair of distinct values (order does not matter for the total):

$$0 + 0 = 0, \quad 0 + 1 = 1, \quad 1 + 1 = 2, \quad 0 + 6 = 6, \quad 1 + 6 = 7, \quad 6 + 6 = 12$$

The possible scores are

$$\{0, \ 1, \ 2, \ 6, \ 7, \ 12\}$$

— six different scores.

The printed answer contradicts itself here. The key reads “$5$ scores $(0, 1, 2, 6, 7, 12)$” — the count says five, and the list beside it has six entries. Counting the list, or building it from the three face values as above, gives six, and that is the correct answer. Three distinct values give $\binom{3}{2} + 3 = 6$ possible totals in general, and none of these six coincide.

(ii) A total of $7$. The only way to reach $7$ is $1 + 6$. Count the ordered pairs:

• first die $1$ (three faces) and second die $6$ (two faces): $3 \times 2 = 6$
• first die $6$ (two faces) and second die $1$ (three faces): $2 \times 3 = 6$

$$6 + 6 = 12 \text{ favourable outcomes}$$

$$P = \frac{12}{36} = \frac13$$

The multiplicities are the whole calculation. Treating the dice as though each value appeared once would give $2$ favourable outcomes out of $9$, and both numbers would be wrong. The faces, not the values, are what is equally likely.

Question 40

A lot consists of $48$ mobile phones of which $42$ are good, $3$ have only minor defects and $3$ have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is (i) acceptable to Varnika? (ii) acceptable to the trader?

(Exercise 13.3, Question 40.)

Solution. The two buyers have different standards, and that is the question.

(i) Acceptable to Varnika. She buys only good phones — not the minor-defect ones, not the major-defect ones:

$$P = \frac{42}{48} = \frac78$$

(ii) Acceptable to the trader. He accepts anything with no major defect, so the good phones and the minor-defect ones:

$$42 + 3 = 45$$

$$P = \frac{45}{48} = \frac{15}{16}$$

The minor-defect phones are the pivot. They fall outside Varnika’s standard and inside the trader’s, which is the only difference between the two answers — three phones, and the probability rises from $\frac78 = \frac{42}{48}$ to $\frac{45}{48}$.

A check on the counts: $42 + 3 + 3 = 48$ ✓, so the three categories are exhaustive.

Reading “no major defect” as “no defect at all” would collapse (ii) into (i) and lose the point of the question.

Question 41

A bag contains $24$ balls of which $x$ are red, $2x$ are white and $3x$ are blue. A ball is selected at random. What is the probability that it is (i) not red? (ii) white?

(Exercise 13.3, Question 41.)

Solution. Find $x$ first. The three colours account for all the balls:

$$x + 2x + 3x = 24 \implies 6x = 24 \implies x = 4$$

So the bag holds

$$4 \text{ red}, \qquad 8 \text{ white}, \qquad 12 \text{ blue}$$

and $4 + 8 + 12 = 24$ ✓.

(i) Not red. By the complement:

$$P = 1 – \frac{4}{24} = \frac{20}{24} = \frac56$$

(ii) White:

$$P = \frac{8}{24} = \frac13$$

Solving for $x$ is the only real step. Attempting the probabilities in terms of $x$ — writing $P(\text{white}) = \frac{2x}{24}$ and stopping — leaves an answer that still contains an unknown, and the question expects a number. The total of $24$ is given precisely so that $x$ can be pinned down.

Note the proportions come out as $1 : 2 : 3$, so blue is half the bag and red a sixth, whatever the total had been.

Question 42

At a fete, cards bearing numbers $1$ to $1000$, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than $500$, the player wins a prize. What is the probability that (i) the first player wins a prize? (ii) the second player wins a prize, if the first has won?

(Exercise 13.3, Question 42.)

Solution. Find the winning cards. We need perfect squares $n^2$ with

$$500 < n^2 \le 1000$$

Since $22^2 = 484$ (too small) and $23^2 = 529$ (the first above $500$), while $31^2 = 961$ (within range) and $32^2 = 1024$ (past $1000$), the winning numbers are

$$529, \ 576, \ 625, \ 676, \ 729, \ 784, \ 841, \ 900, \ 961$$

— the squares of $23$ through $31$, which is nine cards.

(i) The first player wins.

$$P = \frac{9}{1000} = 0.009$$

(ii) The second player wins, given the first has won. The first player has taken a card and it was not replaced — and we are told it was a winning one. So update both counts:

• cards remaining: $1000 – 1 = 999$
• winning cards remaining: $9 – 1 = 8$

$$P = \frac{8}{999}$$

Both numbers must change, exactly as in Question 36. The card removed was a winner, so it leaves the winning pile as well as the box. Reducing only the total would give $\frac{9}{999}$; reducing only the winners would give $\frac{8}{1000}$.

Getting the range right is the other half. Both bounds are strict in different directions — “greater than $500$” excludes $500$ itself (not a square anyway), and the cards stop at $1000$, so $32^2$ is out. Missing $961$ at the top, or including $484$ at the bottom, changes both answers.

Common mistakes

• Questions 19, 20, 21, 26 — collapsing the sample space for two dice. There are $36$ ordered pairs. Counting $(1,3)$ without $(3,1)$ halves the answer in Question 26, and treating the $21$ unordered pairs as equally likely breaks all four.
• Question 39 — ignoring repeated faces. The die shows $0, 1, 1, 1, 6, 6$, so a $1$ is three times as likely as a $0$. Count faces, not values: a total of $7$ has $12$ favourable outcomes, not $2$.
• Questions 28, 29, 30, 36, 42 — leaving the denominator at its original value. Removing cards, bulbs or a drawn card changes the total, and when the item removed was itself favourable, it changes the numerator too.
• Question 31 — reading “between $0$ and $100$” as inclusive. It means $1$ to $99$, ninety-nine integers. The key’s $\frac{14}{99}$ confirms it.
• Question 32 — miscounting the squares. From $2$ to $101$ the squares are $4$ through $100$, nine of them; $1$ is below the range and $121$ above it. Check both ends.
• Question 22 — including the boundary. “Less than $9$” excludes $3 \times 3 = 9$. That single pair is the whole difference between $\frac{16}{36}$ and $\frac{17}{36}$.
• Question 38 — guessing what “otherwise” covers. The rules name one, two and three heads; everything else is the single sequence $TTT$. Work out the leftover case rather than assuming it is large.
• Questions 34 and 35 — weighting by the money. The prize amounts and the slip values never enter. Only the counts matter, because each envelope or slip is equally likely to be drawn.

Practise next

• Exercise 13.3, Questions 1 to 18 — the grouped-data half of this exercise, on mean, median, mode and cumulative frequency tables.
• Exercise 13.2 — worth revisiting now, especially its Questions 5, 9 and 10, which explain why the sample spaces here have to be kept expanded.
• Exercise 13.1 — Questions 18 to 26 are these same counting arguments set as multiple choice, and make a quick final check.