NCERT Class 10 Mathematics — Probability, Exercise 14.1. All 25 questions solved.

This is the only exercise in the chapter, and it runs on one definition. When all the outcomes of an experiment are equally likely, the theoretical probability of an event $\mathrm E$ is

$$\mathrm P(\mathrm E) = \frac{\text{Number of outcomes favourable to } \mathrm E}{\text{Number of all possible outcomes of the experiment}}$$

Three facts follow from it and are used throughout:

$$0 \le \mathrm P(\mathrm E) \le 1, \qquad \mathrm P(\mathrm E) + \mathrm P(\text{not } \mathrm E) = 1$$

An event that is certain has probability $1$ (a sure event); one that cannot happen has probability $0$ (an impossible event).

Key insight. The formula counts outcomes, so it is only valid when every outcome being counted is equally likely. Listing outcomes is easy; making sure they are equally likely is the real work. Questions 2, 22(ii) and 25 test exactly this, and the cure is always the same: make the objects distinguishable — a first coin and a second coin, a blue die and a grey die — so that each listed outcome genuinely has the same chance.

Question 1

Complete the following statements:

(i) Probability of an event $\mathrm E$ + Probability of the event ‘not $\mathrm E$’ $=$ ______.

(ii) The probability of an event that cannot happen is ______. Such an event is called ______.

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

(iv) The sum of the probabilities of all the elementary events of an experiment is ______.

(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.

Solution. $\mathrm E$ and ‘not $\mathrm E$’ between them cover every outcome exactly once, so their favourable counts add up to the total and their probabilities to $1$. An event that cannot happen has no favourable outcomes, so its probability is $\tfrac{0}{n} = 0$; a certain event has all $n$, so $\tfrac nn = 1$. The elementary events (single outcomes) together make up the whole experiment, so their probabilities also sum to $1$. And since the favourable count is never negative and never exceeds the total, every probability lies between $0$ and $1$.

Question 2

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

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

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

Solution. Two outcomes are not automatically equally likely — the question is whether anything favours one over the other.

(i) Not equally likely. Whether the car starts depends on its condition, fuel and battery; a well-maintained car nearly always starts.

(ii) Not equally likely. The result depends on the player’s skill — a good player scores far more often than she misses.

(iii) Equally likely. Answering a true-false question as a trial, with nothing to favour either choice, gives right and wrong the same chance.

(iv) Equally likely. A baby is taken to be as likely to be a boy as a girl — this is the assumption the book makes, and it is very close to the real proportion.

Question 3

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution. A fair decision is one that gives neither team an advantage. When a coin is tossed, head and tail are equally likely, each with probability $\tfrac12$, and the result of a single toss cannot be predicted or influenced. Whichever face a team calls, its chance of winning the toss is the same as the other team’s.

Question 4

Which of the following cannot be the probability of an event?

Solution. A probability must lie between $0$ and $1$. Options (A) and (D) clearly do, and (C) is just another way of writing $0.15$. Only $-1.5$ falls outside — no event can have a negative number of favourable outcomes.

Question 5

If $\mathrm P(\mathrm E) = 0.05$, what is the probability of ‘not $\mathrm E$’?

Solution. $\mathrm E$ and ‘not $\mathrm E$’ are complementary, so their probabilities add to $1$:

$$\mathrm P(\text{not } \mathrm E) = 1 – 0.05 = 0.95$$

Question 6

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

Solution. (i) There are no orange candies, so no outcome is favourable. This is an impossible event: $\mathrm P = 0$.

(ii) Every candy is lemon, so every outcome is favourable. This is a sure event: $\mathrm P = 1$.

Question 7

It is given that in a group of $3$ students, the probability of $2$ students not having the same birthday is $0.992$. What is the probability that the $2$ students have the same birthday?

Solution. “Having the same birthday” and “not having the same birthday” are complements of each other, so

$$\mathrm P(\text{same birthday}) = 1 – 0.992 = 0.008$$

Question 8

A bag contains $3$ red balls and $5$ black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Solution. “At random” means each of the $3 + 5 = 8$ balls is equally likely to be drawn.

$$\text{(i)}\ \ \mathrm P(\text{red}) = \frac{3}{8}, \qquad \text{(ii)}\ \ \mathrm P(\text{not red}) = 1 – \frac{3}{8} = \frac{5}{8}$$

The second agrees with counting directly: the $5$ black balls are the ones that are not red.

Question 9

A box contains $5$ red marbles, $8$ white marbles and $4$ green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

Solution. There are $5 + 8 + 4 = 17$ equally likely marbles.

$$\text{(i)}\ \ \frac{5}{17}, \qquad \text{(ii)}\ \ \frac{8}{17}, \qquad \text{(iii)}\ \ 1 – \frac{4}{17} = \frac{13}{17}$$

For (iii), the $5 + 8 = 13$ red and white marbles are exactly the ones that are not green, which confirms it.

Question 10

A piggy bank contains hundred $50$p coins, fifty ₹ $1$ coins, twenty ₹ $2$ coins and ten ₹ $5$ coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a $50$ p coin? (ii) will not be a ₹ $5$ coin?

Solution. The equally likely outcomes are the individual coins, not the four kinds of coin: there are $100 + 50 + 20 + 10 = 180$ of them.

(i) $100$ of the $180$ coins are $50$p coins: $\mathrm P = \dfrac{100}{180} = \dfrac{5}{9}$.

(ii) $10$ coins are ₹ $5$ coins, so $180 – 10 = 170$ are not: $\mathrm P = \dfrac{170}{180} = \dfrac{17}{18}$.

Question 11

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing $5$ male fish and $8$ female fish (see Fig. 14.4). What is the probability that the fish taken out is a male fish?

Solution. Fig. 14.4 shows the shopkeeper netting a fish from the tank; it adds nothing to the numbers. Each of the $5 + 8 = 13$ fish is equally likely to be caught, and $5$ are male.

$$\mathrm P(\text{male}) = \frac{5}{13}$$

Question 12

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8$ (see Fig. 14.5), and these are equally likely outcomes. What is the probability that it will point at

(i) $8$?

(ii) an odd number?

(iii) a number greater than $2$?

(iv) a number less than $9$?

Solution. Fig. 14.5 is a circle divided into eight equal sectors numbered $1$ to $8$, with an arrow pivoted at the centre. There are $8$ equally likely outcomes.

(i) Only one sector shows $8$: $\tfrac18$.

(ii) The odd numbers are $1, 3, 5, 7$ — four of them: $\tfrac48 = \tfrac12$.

(iii) “Greater than $2$” means $3, 4, 5, 6, 7, 8$ — six of them: $\tfrac68 = \tfrac34$.

(iv) Every number on the spinner is less than $9$, so this is a sure event: $\tfrac88 = 1$.

Question 13

A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between $2$ and $6$;

(iii) an odd number.

Solution. The six faces $1, 2, 3, 4, 5, 6$ are equally likely.

(i) The primes are $2, 3, 5$. Note that $1$ is not prime: $\tfrac36 = \tfrac12$.

(ii) “Between $2$ and $6$” excludes both ends, leaving $3, 4, 5$: $\tfrac36 = \tfrac12$.

(iii) The odd faces are $1, 3, 5$: $\tfrac36 = \tfrac12$.

Question 14

One card is drawn from a well-shuffled deck of $52$ cards. Find the probability of getting

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Solution. A deck has four suits of $13$ cards — spades and clubs are black, hearts and diamonds red. The face cards are the jack, queen and king of each suit, so there are $12$ of them. Well-shuffling makes all $52$ cards equally likely.

(i) Red kings: the king of hearts and the king of diamonds, $2$ cards: $\tfrac{2}{52} = \tfrac{1}{26}$.

(ii) $\tfrac{12}{52} = \tfrac{3}{13}$.

(iii) Half the face cards are red, $6$ cards: $\tfrac{6}{52} = \tfrac{3}{26}$.

(iv) Exactly one card: $\tfrac{1}{52}$.

(v) $13$ spades: $\tfrac{13}{52} = \tfrac14$.

(vi) Exactly one card: $\tfrac{1}{52}$.

Question 15

Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Solution. (i) Five equally likely cards, one queen: $\tfrac15$.

(ii) Once the queen is put aside, the experiment changes: only the ten, jack, king and ace remain, so there are $4$ equally likely outcomes.

(a) One ace among four cards: $\tfrac14$.

(b) There is no queen left, so this is impossible: $0$.

Question 16

$12$ defective pens are accidentally mixed with $132$ good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution. Since the pens cannot be told apart by looking, the draw is genuinely random: each of the $12 + 132 = 144$ pens is equally likely.

$$\mathrm P(\text{good}) = \frac{132}{144} = \frac{11}{12}$$

Question 17

(i) A lot of $20$ bulbs contain $4$ defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution. (i) $\tfrac{4}{20} = \tfrac15$.

(ii) A good bulb has been removed and not put back, so the lot now has $19$ bulbs: still $4$ defective, but only $16 – 1 = 15$ good. Both the total and the favourable count have changed.

$$\mathrm P(\text{not defective}) = \frac{15}{19}$$

Question 18

A box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by $5$.

Solution. There are $90$ equally likely discs.

(i) The one-digit numbers are $1$ to $9$, so the two-digit ones are $10$ to $90$: $90 – 9 = 81$ of them. $\tfrac{81}{90} = \tfrac{9}{10}$.

(ii) The perfect squares up to $90$ are $1, 4, 9, 16, 25, 36, 49, 64, 81$ — nine of them, since $10^2 = 100$ is too big. $\tfrac{9}{90} = \tfrac{1}{10}$.

(iii) The multiples of $5$ are $5, 10, \ldots, 90$: $\tfrac{90}{5} = 18$ of them. $\tfrac{18}{90} = \tfrac15$.

Question 19

A child has a die whose six faces show the letters as given below:

$$\boxed{\mathrm A}\quad \boxed{\mathrm B}\quad \boxed{\mathrm C}\quad \boxed{\mathrm D}\quad \boxed{\mathrm E}\quad \boxed{\mathrm A}$$

The die is thrown once. What is the probability of getting (i) A? (ii) D?

Solution. The equally likely outcomes are the six faces, not the five letters. The letter A is on two faces.

$$\text{(i)}\ \ \mathrm P(\mathrm A) = \frac{2}{6} = \frac13, \qquad \text{(ii)}\ \ \mathrm P(\mathrm D) = \frac16$$

Question 20

20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter $1$ m?

The figure is a rectangle $3$ m long and $2$ m wide with a circle of diameter $1$ m drawn inside it.

The asterisk. The book marks this question with an asterisk, and its footnote says the question is not from the examination point of view. The chapter uses the same marker on its Examples 10 and 11, which introduce probability measured by lengths and areas rather than by counting.

Solution. The die can land at any point of the rectangle, and there are infinitely many points, so we cannot count outcomes. Following Examples 10 and 11, we compare areas instead: every point is equally likely, so the chance of landing in the circle is the fraction of the rectangle’s area that the circle covers.

$$\text{Area of rectangle} = 3 \times 2 = 6\ \text{m}^2$$

$$\text{Area of circle} = \pi\left(\tfrac12\right)^2 = \frac{\pi}{4}\ \text{m}^2$$

$$\mathrm P(\text{lands inside the circle}) = \frac{\pi/4}{6} = \frac{\pi}{24} \approx 0.131$$

Question 21

A lot consists of $144$ ball pens of which $20$ are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

Solution. “She will buy it” is the same event as “the pen is good”, and there are $144 – 20 = 124$ good pens among $144$ equally likely ones.

$$\text{(i)}\ \ \frac{124}{144} = \frac{31}{36}, \qquad \text{(ii)}\ \ \frac{20}{144} = \frac{5}{36}$$

The two are complementary, and $\tfrac{31}{36} + \tfrac{5}{36} = 1$.

Question 22

Refer to Example 13. (i) Complete the following table:

Event: ‘Sum on 2 dice’	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$						$\frac{5}{36}$				$\frac{1}{36}$

(ii) A student argues that ‘there are $11$ possible outcomes $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ and $12$. Therefore, each of them has a probability $\frac{1}{11}$’. Do you agree with this argument? Justify your answer.

Solution. Example 13 throws a blue die and a grey die together. Because the dice are distinguishable, $(1, 4)$ and $(4, 1)$ are different outcomes, and there are $6 \times 6 = 36$ equally likely ordered pairs.

(i) Count the pairs giving each sum:

Sum	Favourable pairs	Count
$2$	$(1,1)$	$1$
$3$	$(1,2), (2,1)$	$2$
$4$	$(1,3), (2,2), (3,1)$	$3$
$5$	$(1,4), (2,3), (3,2), (4,1)$	$4$
$6$	$(1,5), (2,4), (3,3), (4,2), (5,1)$	$5$
$7$	$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$	$6$
$8$	$(2,6), (3,5), (4,4), (5,3), (6,2)$	$5$
$9$	$(3,6), (4,5), (5,4), (6,3)$	$4$
$10$	$(4,6), (5,5), (6,4)$	$3$
$11$	$(5,6), (6,5)$	$2$
$12$	$(6,6)$	$1$

The counts add to $36$, so every pair has been counted exactly once. Dividing each count by $36$ completes the table:

Event: ‘Sum on 2 dice’	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

The two entries the book gives, $\tfrac{5}{36}$ for a sum of $8$ and $\tfrac{1}{36}$ for $12$, agree.

(ii) No. There are indeed eleven possible sums, but they are not equally likely, so the formula cannot be applied to them. A sum of $7$ can happen in six ways and a sum of $2$ in only one, so $7$ is six times as likely as $2$.

Question 23

A game consists of tossing a one rupee coin $3$ times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution. Each toss has $2$ equally likely results, so three tosses give $2 \times 2 \times 2 = 8$ equally likely sequences:

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

where, for example, THH means a tail on the first toss and heads on the second and third. Hanif wins only on HHH and TTT, so he loses on the other $6$.

$$\mathrm P(\text{lose}) = \frac{6}{8} = \frac34$$

Question 24

A die is thrown twice. What is the probability that

(i) $5$ will not come up either time?

(ii) $5$ will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution. By the hint, this is the $36$-outcome experiment of Example 13, with “first throw” and “second throw” in place of the blue and grey dice.

(i) For $5$ never to appear, each throw must show one of the other $5$ faces, so there are $5 \times 5 = 25$ favourable pairs.

$$\mathrm P(\text{no } 5) = \frac{25}{36}$$

(ii) “At least once” is the complement of “not at all”:

$$\mathrm P(\text{at least one } 5) = 1 – \frac{25}{36} = \frac{11}{36}$$

Counting directly agrees: $6$ pairs have $5$ first, $6$ have $5$ second, and $(5, 5)$ is in both lists, so $6 + 6 – 1 = 11$.

Question 25

Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\frac13$.

(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is $\frac12$.

Solution. (i) Incorrect. Call the coins first and second. The equally likely outcomes are HH, HT, TH and TT. “One of each” happens in two of these ways — head on the first coin and tail on the second, or the other way round — so it is twice as likely as two heads or two tails:

$$\mathrm P(\text{two heads}) = \tfrac14, \qquad \mathrm P(\text{two tails}) = \tfrac14, \qquad \mathrm P(\text{one of each}) = \tfrac24 = \tfrac12$$

The three outcomes can be listed, but they are not equally likely.

(ii) Correct. Here the two outcomes really are equally likely: three faces ($1, 3, 5$) are odd and three ($2, 4, 6$) are even, so each has probability $\tfrac36 = \tfrac12$.

Common mistakes

• Question 4, rejecting $15\%$. A percentage is just a fraction out of $100$: $15\% = 0.15$, which is a perfectly good probability. The only option outside $[0, 1]$ is $-1.5$.
• Question 10, counting the kinds of coin. There are four kinds, but they are not equally likely to fall out — there are ten times as many $50$p coins as ₹ $5$ coins. The equally likely outcomes are the $180$ individual coins.
• Question 14(ii), getting the number of face cards wrong. Face cards are the jack, queen and king — $3$ per suit, $12$ in all. Counting the ace, or only one suit, gives $\tfrac{16}{52}$ or $\tfrac{3}{52}$.
• Question 17(ii), changing only one number. A good bulb has gone, so both the total ($19$) and the good bulbs ($15$) drop by one. Writing $\tfrac{16}{19}$ or $\tfrac{15}{20}$ changes only one of them.
• Question 18(i), miscounting $10$ to $90$. The count of whole numbers from $10$ to $90$ inclusive is $90 – 10 + 1 = 81$, not $80$.
• Question 20, using the diameter as the radius. The circle’s diameter is $1$ m, so its radius is $\tfrac12$ m and its area $\tfrac{\pi}{4}$ m², not $\pi$ m².
• Question 24(ii), adding $\tfrac16 + \tfrac16$. That counts $(5, 5)$ twice and gives $\tfrac{12}{36}$. Going through the complement, $1 – \tfrac{25}{36}$, avoids the double count altogether.
• Questions 22(ii) and 25(i), assuming every listed outcome is equally likely. This is the error the whole exercise is built to expose. Before dividing by the number of outcomes, ask whether each can happen in the same number of ways.

Practise next

• Class 11 Probability, Exercise 14.1 — events described as sets, where “not $\mathrm E$”, “$\mathrm A$ or $\mathrm B$” and “$\mathrm A$ and $\mathrm B$” become complement, union and intersection.
• Class 11 Probability, Exercise 14.2 — the axiomatic approach, which starts from the same facts used here, $0 \le \mathrm P(\mathrm E) \le 1$ and $\mathrm P(\text{not } \mathrm E) = 1 – \mathrm P(\mathrm E)$.