NCERT Solutions Class 9 Maths Chapter 7 End-of-Chapter Exercises — The Mathematics of Maybe: Introduction to Probability

NCERT Class 9 Mathematics — Ganita Manjari Part I, The Mathematics of Maybe: Introduction to Probability, End-of-Chapter Exercises. All 16 questions solved.

Everything from the chapter comes back here: the $0$-to-$1$ scale, experimental probability from a data table, theoretical probability from a sample space, tree diagrams with and without replacement, and — in the last question — a probability measured by area rather than by counting.

$$P(E) = \frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}, \qquad 0 \le P(E) \le 1, \qquad P(\text{not } E) = 1 – P(E)$$

Key insight. That formula is only valid when the outcomes are equally likely, and half the questions here are really testing whether you have noticed. Counting favourable leaves out of total leaves gives the wrong answer in Questions 3, 4(iv) and 12(ii) unless you first choose a sample space in which every outcome carries the same weight.

Questions $9$ to $16$ carry an asterisk in the book, which marks them as not assessed. They are the most interesting ones on the page.

Question 1

Fill in the blanks.

(i) The probability of an impossible event is ______

Solution. An impossible event has no favourable outcomes at all, so the numerator of $\frac{\text{favourable}}{\text{possible}}$ is $0$. The answer is $0$.

(ii) The set of all possible outcomes of a random experiment is called the ______

Solution. The sample space, usually written $S$.

(iii) The probability of an event that is certain to happen is ______

Solution. Every outcome is favourable, so the fraction is $\frac{n(S)}{n(S)} = 1$. The answer is $1$.

(iv) Tossing a fair coin has a probability of ______ for getting heads

Solution. Two equally likely outcomes, one of them favourable: $\frac12$ (or $0.5$, or $50\%$).

Question 2

In a survey of $50$ students, $15$ students said they liked football. The number of students who like football is $15$, and the ______ (frequency/relative frequency) is ______ (fill in the fraction or decimal).

Solution. The raw count $15$ is the frequency. What the sentence goes on to ask for is therefore the relative frequency — the frequency divided by the total number surveyed:

$$\text{relative frequency} = \frac{15}{50} = \frac{3}{10} = 0.3$$

Relative frequency is experimental probability under another name: it is the proportion of the sample in which the event occurred, and it is the estimate we would use for the probability that a randomly chosen student likes football.

Question 3

Which of the following experiments have equally likely outcomes? Explain.

Question 3 (i)

A driver attempts to start a car. The car starts or does not start.

Solution. Not equally likely. There are two outcomes, but nothing makes them balanced. A car in good repair starts almost every time, so “starts” is far more likely than “does not start”; a car with a dying battery is the other way round. Two outcomes never means a probability of $\frac12$ each on their own — some symmetry has to force it, and here there is none.

Question 3 (ii)

Tossing a fair coin once.

Solution. Equally likely. The word fair is doing the work: it means precisely that the two faces are equally likely, so $P(H) = P(T) = \frac12$.

Question 3 (iii)

Rolling a fair $6$-sided die.

Solution. Equally likely. A fair die is symmetric — no face is favoured over another — so each of $1, 2, 3, 4, 5, 6$ has probability $\frac16$.

Question 3 (iv)

Choosing a marble randomly from a bag that contains $3$ red marbles and $7$ blue marbles.

Solution. It depends on what you call an outcome, and this is the most instructive part of the question.

If the outcome recorded is the colour, the two outcomes are not equally likely:

$$P(\text{red}) = \frac{3}{10}, \qquad P(\text{blue}) = \frac{7}{10}$$

If the outcome recorded is which marble was drawn, then all ten marbles are equally likely, each with probability $\frac{1}{10}$ — and that is where the two figures above come from.

As the question is normally read — red or blue — the answer is not equally likely.

Question 3 (v)

A baby is born. It is a boy or a girl.

Solution. Equally likely, near enough. This is the usual modelling assumption and it is close to true: $P(\text{boy}) = P(\text{girl}) = \frac12$.

It is worth knowing that it is not exactly true. Slightly more boys than girls are born worldwide — roughly $105$ boys to every $100$ girls — so $P(\text{boy})$ is nearer $0.512$ than $0.500$. For a school problem the difference is negligible; the point is that this is an empirical near-equality, not a symmetry like the die’s.

Question 4

Write the sample space and calculate the probability based on the given information.

Question 4 (i)

Two coins are tossed at the same time. What is the probability of getting at least one head?

Solution.

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

“At least one head” means one head or two — everything except $TT$:

$$E = \{HH, HT, TH\}, \qquad P(E) = \frac{3}{4}$$

The quickest route is the complement: $P(\text{at least one head}) = 1 – P(TT) = 1 – \frac14 = \frac34$.

Question 4 (ii)

Ten identical cards numbered $1$ to $10$ are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?

Solution.

$$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, \qquad E = \{2, 4, 6, 8, 10\}$$

$$P(E) = \frac{5}{10} = \frac{1}{2}$$

Question 4 (iii)

A die is rolled once. What is the probability of getting a number greater than $4$?

Solution.

$$S = \{1, 2, 3, 4, 5, 6\}, \qquad E = \{5, 6\}$$

$$P(E) = \frac{2}{6} = \frac{1}{3}$$

Greater than $4$ excludes $4$ itself. “At least $4$” would have given $\frac36 = \frac12$.

Question 4 (iv)

A bag contains $3$ red balls, $2$ blue balls, and $1$ green ball. One ball is picked at random. What is the probability that it is not red?

Solution. There are $3 + 2 + 1 = 6$ balls, and taking the balls themselves as the outcomes makes them equally likely. Not red means blue or green — $2 + 1 = 3$ balls:

$$P(\text{not red}) = \frac{3}{6} = \frac{1}{2}$$

Or by the complement: $P(\text{red}) = \frac36 = \frac12$, so $P(\text{not red}) = 1 – \frac12 = \frac12$.

Question 4 (v)

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Solution.

$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}, \qquad n(S) = 8$$

Exactly two heads — no more, no fewer:

$$E = \{HHT, HTH, THH\}, \qquad P(E) = \frac{3}{8}$$

$HHH$ does not qualify: it has three heads, not exactly two.

Question 5

A bag has $3$ candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Solution. Three candies, one of each kind, picked at random — so the three outcomes are equally likely:

$$S = \{\text{strawberry}, \text{lemon}, \text{mint}\}, \qquad P(\text{strawberry}) = \frac{1}{3}$$

Question 6

A child has $2$ shirts (one red and one blue) and $3$ types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

Solution. Two choices of shirt and three of pants give $2 \times 3 = 6$ outfits. A table with shirts down the side and pants across the top lists them all, with no risk of repeating or skipping one:

	Jeans	Khakis	Shorts
Red shirt	Red, jeans	Red, khakis	Red, shorts
Blue shirt	Blue, jeans	Blue, khakis	Blue, shorts

Written as a sample space:

$$S = \{(R, J), (R, K), (R, S), (B, J), (B, K), (B, S)\}, \qquad n(S) = 6$$

The table is the multiplication principle made visible: $2$ rows $\times$ $3$ columns $= 6$ cells, one per outfit.

Question 7

A tyre company records distances before replacement in $1000$ cases.

Distance (km)	Less than $4000$	$4001$ to $9000$	$9001$ to $14000$	More than $14000$
Number of cases	$20$	$210$	$325$	$445$

Find the probability that a randomly chosen tyre lasts:

Solution. These are experimental probabilities, read straight off the data: each is a frequency divided by the $1000$ cases. (Check first that the table is complete: $20 + 210 + 325 + 445 = 1000$ ✓)

(i) Less than $4000$ km

$$P = \frac{20}{1000} = \frac{1}{50} = 0.02$$

(ii) Between $4000$ and $14000$ km

Two bands fall in this range — $4001$ to $9000$ and $9001$ to $14000$:

$$P = \frac{210 + 325}{1000} = \frac{535}{1000} = \frac{107}{200} = 0.535$$

(iii) More than $14000$ km

$$P = \frac{445}{1000} = \frac{89}{200} = 0.445$$

As a check the three probabilities cover every case exactly once, so they must total $1$:

$$0.02 + 0.535 + 0.445 = 1 \ \checkmark$$

Question 8

The letters of the word ‘PEACE’ are placed on cards — P, E, A, C, E. Leela draws a card without looking.

Solution. Five cards, drawn blind, so all five are equally likely. Note that E appears twice: the multiset of letters is $\{P, E, A, C, E\}$, not $\{P, E, A, C\}$.

(i) What is the probability that it is a P, E or C?

Favourable cards: the P, both E’s, and the C — that is $1 + 2 + 1 = 4$ cards.

$$P = \frac{4}{5} = 0.8$$

(ii) What is the probability that it is not an E?

Three cards are not an E — P, A and C:

$$P(\text{not E}) = \frac{3}{5} = 0.6$$

Equivalently $P(E) = \frac25$, so $P(\text{not E}) = 1 – \frac25 = \frac35$.

Question 9

A game of chance consists of spinning an arrow (see Fig. 7.7) which comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8$, and these are equally likely outcomes. What is the probability that it will point at

Solution. The question hands us the crucial fact: the eight sectors are equally likely, so each has probability $\frac18$ and every part is a matter of counting the favourable sectors.

$$S = \{1, 2, 3, 4, 5, 6, 7, 8\}, \qquad n(S) = 8$$

(i) $8$?

One sector out of eight:

$$P = \frac{1}{8}$$

(ii) An odd number?

$E = \{1, 3, 5, 7\}$ — four sectors:

$$P = \frac{4}{8} = \frac{1}{2}$$

(iii) A number greater than $2$?

$E = \{3, 4, 5, 6, 7, 8\}$ — six sectors:

$$P = \frac{6}{8} = \frac{3}{4}$$

(iv) A number less than $9$?

Every sector is less than $9$, so this is a certain event:

$$P = \frac{8}{8} = 1$$

(v) A multiple of $3$?

$E = \{3, 6\}$ — note $9$ is not on the spinner:

$$P = \frac{2}{8} = \frac{1}{4}$$

Question 10

A basket contains $4$ red balls and $5$ blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.

Solution. The first ball is laid aside, so this is drawing without replacement: the second stage has only $8$ balls left, and which ones depends on what the first draw took. That is what makes the two levels of the tree different.

The four leaves add to $\frac{12 + 20 + 20 + 20}{72} = \frac{72}{72} = 1$ ✓

(i) What is the probability of drawing a red ball and then a blue ball?

Follow the red branch and then the blue one. The first draw is red with probability $\frac49$; that leaves $3$ red and $5$ blue, so the second is blue with probability $\frac58$:

$$P(RB) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18}$$

(ii) What is the probability of drawing $2$ blue balls?

$$P(BB) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18}$$

The two answers coming out equal is a coincidence of these particular numbers, not a rule — the paths are quite different. It is worth noticing, though, because it is a useful reminder that equal answers do not mean you have made a mistake.

Question 11

I throw a pair of $6$-sided dice. Write down an event that has a probability of $0$ and an outcome that has a probability of $1$.

Solution. With two dice the sample space has $6 \times 6 = 36$ ordered pairs, and the possible totals run from $2$ to $12$.

An event of probability $0$ — anything that cannot happen. For example:

• “The sum is $1$.” The smallest possible sum is $1 + 1 = 2$, so no outcome qualifies: $P = \frac{0}{36} = 0$.
• “The sum is $13$”, or “both dice show $7$”, would do equally well.

An event of probability $1$ — anything certain. For example:

• “The sum lies between $2$ and $12$ inclusive”: all $36$ outcomes qualify, so $P = \frac{36}{36} = 1$.
• “Each die shows a whole number from $1$ to $6$” is the same idea stated differently.

A note on the wording. The question asks for an outcome of probability $1$, and strictly there is no such thing here: every individual outcome of a pair of dice is one of the $36$ pairs, each with probability $\frac{1}{36}$. An outcome could only have probability $1$ if the experiment had just one possible result. What is meant is a certain event — an event containing the whole sample space — and that is what is given above.

Question 12

Write the sample space and calculate the probability based on the given information.

Question 12 (i)

Two dice are rolled. What is the probability that the sum is a prime number greater than $5$?

Solution. The sample space is the $36$ ordered pairs $(a, b)$ with $a, b \in \{1, \dots, 6\}$, all equally likely.

Sums range from $2$ to $12$. The primes in that range are $2, 3, 5, 7, 11$, and those greater than $5$ are $7$ and $11$ only. ($9$ is not prime, since $9 = 3 \times 3$.)

• Sum $7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — $6$ outcomes.
• Sum $11$: $(5,6), (6,5)$ — $2$ outcomes.

$$P = \frac{6 + 2}{36} = \frac{8}{36} = \frac{2}{9}$$

Question 12 (ii)

A bag contains $4$ red, $3$ green, and $2$ blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

Solution. There are $9$ balls, and two are drawn together, so the sample space is the set of unordered pairs:

$$n(S) = \binom{9}{2} = \frac{9 \times 8}{2} = 36$$

It is far easier to count the same-colour pairs and subtract:

$$\binom{4}{2} + \binom{3}{2} + \binom{2}{2} = 6 + 3 + 1 = 10$$

so different colours accounts for $36 – 10 = 26$ pairs:

$$P(\text{different colours}) = \frac{26}{36} = \frac{13}{18} \approx 0.722$$

Counted directly, the mixed pairs are $4 \times 3 = 12$ red–green, $4 \times 2 = 8$ red–blue and $3 \times 2 = 6$ green–blue, totalling $26$ ✓

Question 12 (iii)

Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

Solution. From the eight equally likely outcomes

$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

we need both conditions: the first letter is $H$, and there are exactly two $H$’s altogether.

• $HHT$ ✓ — starts with $H$, two heads.
• $HTH$ ✓ — starts with $H$, two heads.
• $THH$ ✗ — two heads, but starts with $T$.
• $HHH$ ✗ — starts with $H$, but three heads.

$$P = \frac{2}{8} = \frac{1}{4}$$

Question 12 (iv)

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. Using all four digits once each gives $4! = 24$ numbers, all equally likely.

A number is even exactly when its last digit is even — here, $2$ or $4$. Fix the last digit as one of those two, and the remaining three digits can be arranged in $3! = 6$ ways:

$$\text{favourable} = 2 \times 3! = 2 \times 6 = 12$$

$$P(\text{even}) = \frac{12}{24} = \frac{1}{2}$$

There is a quicker way to see it: by symmetry each of the four digits is equally likely to land in the units place, and two of the four are even — so the probability is $\frac24 = \frac12$ without any counting at all.

Question 12 (v)

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. Guessing gives each question a probability $\frac14$ of being right and $\frac34$ of being wrong, independently. The sample space of answer patterns has $4^3 = 64$ equally likely members.

Choose which two of the three questions are the correct ones — $\binom{3}{2} = 3$ ways. For each such choice: the two correct answers are forced (one way each), and the remaining question must be wrong, which can happen in $3$ ways. So

$$\text{favourable} = 3 \times 1 \times 1 \times 3 = 9$$

$$P(\text{exactly } 2 \text{ correct}) = \frac{9}{64} \approx 0.141$$

Equivalently, in probability form:

$$\binom{3}{2}\left(\frac14\right)^{2}\left(\frac34\right) = 3 \times \frac{1}{16} \times \frac34 = \frac{9}{64}$$

The “$\times 3$” at the end matters: getting the third question wrong is not one outcome but three, because there are three wrong options to pick.

Question 13

A box contains $4$ balls numbered $1$ to $4$. Record a sample space using a tree diagram for the following experiments:

Solution. Both experiments have two stages of four-or-fewer branches. The tree has four branches at the first level in each case; what changes is the second level.

(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded

With replacement. Every one of the four first-stage branches splits into the same four branches, because the returned ball is available again — including the possibility of drawing the same ball twice.

$$S_1 = \left\{ \begin{array}{llll} (1,1), & (1,2), & (1,3), & (1,4) \\ (2,1), & (2,2), & (2,3), & (2,4) \\ (3,1), & (3,2), & (3,3), & (3,4) \\ (4,1), & (4,2), & (4,3), & (4,4) \end{array} \right\}$$

That is $4 \times 4 = 16$ outcomes, each of probability $\frac{1}{16}$.

(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball

Without replacement. Each first-stage branch now splits into only three, since the ball already drawn cannot come again. The diagonal pairs $(1,1), (2,2), (3,3), (4,4)$ disappear:

$$S_2 = \left\{ \begin{array}{lll} (1,2), & (1,3), & (1,4) \\ (2,1), & (2,3), & (2,4) \\ (3,1), & (3,2), & (3,4) \\ (4,1), & (4,2), & (4,3) \end{array} \right\}$$

That is $4 \times 3 = 12$ outcomes, each of probability $\frac{1}{12}$.

(iii) What are the sizes of these two sample spaces?

$$n(S_1) = 4 \times 4 = 16, \qquad n(S_2) = 4 \times 3 = 12$$

The difference is exactly the four repeated pairs that replacement allows and removal forbids: $16 – 12 = 4$.

Question 14

List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of $6$ cards numbered $1$ through $6$.

Solution. Each outcome records both results, so it is a pair (coin, card). Two coin results and six cards give $2 \times 6 = 12$ outcomes:

$$S = \{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6),$$ $$\qquad (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\}$$

All twelve are equally likely, each with probability $\frac{1}{12}$. This is the same structure as Exercise 7.3 Question 2(i) — a coin alongside a six-way choice — which is worth noticing: the objects differ but the sample space does not.

Question 15

Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?

Solution. What is recorded is not the pattern $HHT$ but the count of heads. Tossing three coins can give $0$, $1$, $2$ or $3$ heads, and each of those genuinely occurs — $TTT$, $HTT$, $HHT$ and $HHH$ respectively. A sample space must list every possible outcome and nothing else, so:

(i) $\{1, 2, 3\}$

Not a sample space. It omits $0$, and $0$ heads is perfectly possible — it is the outcome $TTT$. A list that leaves out a possible result cannot be a sample space.

(ii) $\{0, 1, 2\}$

Not a sample space. It omits $3$, which happens when all three coins land heads.

(iii) $\{0, 1, 2, 3, 4\}$

Not a sample space. It includes $4$, and four heads is impossible with three coins. A sample space contains only outcomes that can occur; an impossible entry does not belong.

(iv) $\{0, 1, 2, 3\}$

This is the sample space. Every possible number of heads is there, and nothing impossible is.

The two ways a candidate list can fail are exactly the two on display here: it can be incomplete, as (i) and (ii) are, or it can be too generous, as (iii) is. Both matter — an extra entry would drag the denominator up to $5$ and make every probability computed from it wrong.

Question 16

Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8, which measures $3$ m by $2$ m. What is the probability that it will land inside the circle with a diameter of $1$ m?

Solution. Here the outcomes are not countable — the dye can land at any point of the rectangle — so counting has to be replaced by measuring. “At random” means every point is as likely as every other, which makes the probability the ratio of areas:

$$P(\text{event}) = \frac{\text{favourable area}}{\text{total area}}$$

Total area. The rectangle is $3$ m by $2$ m:

$$A_{\text{rect}} = 3 \times 2 = 6 \text{ m}^2$$

Favourable area. The circle has diameter $1$ m, so its radius is $r = \frac12$ m:

$$A_{\text{circle}} = \pi r^2 = \pi \left(\frac12\right)^2 = \frac{\pi}{4} \text{ m}^2$$

The probability.

$$P = \frac{\pi/4}{6} = \frac{\pi}{24} \approx 0.1309$$

Taking $\pi = \frac{22}{7}$ instead gives

$$P = \frac{22}{7 \times 24} = \frac{22}{168} = \frac{11}{84} \approx 0.1310$$

So about a $13\%$ chance — roughly one drop in eight. A sense check: the circle’s area is $0.785$ m$^2$ against the rectangle’s $6$ m$^2$, which is a little over an eighth, and that is what the answer says.

This is called geometric probability, and it is the first time in the chapter that the sample space is infinite. The formula $\frac{\text{favourable}}{\text{possible}}$ still holds — it is just that “how many” has become “how much”.

Common mistakes

• Question 3(i), answering “two outcomes, so $\frac12$ each”. Two outcomes never forces equal probability. A symmetry has to make them equal, and a car starting has none.
• Question 4(v), counting $HHH$ as two heads. “Exactly two” excludes three. Three of the eight outcomes qualify, not four.
• Question 7(ii), using one band instead of two. “Between $4000$ and $14000$” spans the $4001$–$9000$ band and the $9001$–$14000$ band: $210 + 325 = 535$, not $325$.
• Question 8, treating PEACE as four distinct letters. There are five cards and two of them are E. Using a denominator of $4$ gives $\frac34$ and $\frac24$, both wrong.
• Question 9(v), counting $9$ as a multiple of $3$ on the spinner. The spinner stops at $8$. Only $3$ and $6$ are there.
• Question 10, keeping $\frac49$ and $\frac59$ at the second stage. The first ball is laid aside, so the second draw is out of $8$ balls and the probabilities change. Using $9$ twice is the commonest error in the whole chapter.
• Question 12(i), including $9$ as a prime. $9 = 3 \times 3$. The primes above $5$ that two dice can make are $7$ and $11$ only.
• Question 12(v), forgetting the three ways to be wrong. The wrong answer is not one outcome but three, which is why the answer is $\frac{9}{64}$ and not $\frac{3}{64}$.
• Question 16, using the diameter as the radius. $\pi (1)^2 = \pi$ would make the circle bigger than half the rectangle, which the figure plainly contradicts. The radius is $\frac12$ m.

Practise next

• Exercise Set 7.4 — more tree diagrams, all of them with replacement, which is the easier case to draw before returning to Question 10 here.
• Exercise Set 7.2 — revise experimental probability if Questions 2 and 7 gave trouble; they are the same idea with different data.
• Exercise Set 7.3 — listing sample spaces, which is where Questions 13, 14 and 15 come from and the skill the whole page rests on.