# NCERT Solutions for Class 10 Maths Chapter 15 - Probability

Important Points

1. Sample Space : Sample space is the set of all possible outcomes in the experiment. It is denoted by the letters.

If you slip a coin one time, is ‘‘Heads or tails’’ and we write sample space of event as S = {H, T}.

2. Elementary Event : An event which contains only a single outcome in the sample space is called an elementary event.

In tossing a coin, possible.

3. Impossible and Sure Events : If the probability of occurrence of an event is 0, such an event is called an impossible event and if the probability of occurrence of an event is 1, it is called a sure event.

Example : What is the probability of rolling a 7 on six-sided die ? As the 7 never appears on a face of a six sided die, the event is impossible.

The probability of getting a number when a die is thrown is a sure event.

4. Equally Likely Events : Two or more events are equally likely events, if none of them is based over the other.

5. Complementary Event : The complementary event is the set of all sample points of the space other than the sample points of A and $$\text{it is denoted by}\space\bar{\text{A}}\text{of A'.}$$

e.g., Let S = {1, 2, 3, 4, 5, 6}, if A = {1, 2, 4}, then A′ = {3, 5, 6}

6. Probability : Suppose there are n elementary events associated with a random experiment and m of them are favourable to an event A. The probability of happening of occurrence of event

A is defined as the ratio $$\frac{m}{n}$$ and it is denoted by the symbol P(A).

$$\text{i.e.\space}\text{P(A)}=\frac{m}{n}\\=\text{or P(A) =}\frac{\text{Number of favourable outcomes to A}}{\text{Number of total outcomes of A}}$$

Note :

• O ≤ P(A) ≤ 1

• P(A) + P(A′) = 1, where A and A′ are
complementary event.

Remember :

(i) In tossing of a coin, their is only head and tail.

(ii) In tossing of a die, numbers are 1, 2, 3, 4, 5 and 6.

(iii) In a deck of cards, their are 52 cards, which are divided into 4 suits, hearts (♥), diamonds (♦), clubs (♣) spades (♠) of 13 cards each. Total face cards 12, Black face cards 6 and Red face cards 6.

Exercise 15.1

1. Complete the following statements :

(i) Probability of an event E + Probability of the event ‘not 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

Sol. (i) 1, Because the sum of probability of complementary events is equal to one.

(ii) O, Impossible event. This type of events is called impossible events and probability of this event is 0.)

(iii) 1, Sure event, This type of events is called sure events and probability of this event is 1.

(iv) 1, (e.g., suppose a die is tossed, then P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

(v) 0, 1, (since, the probability always lies between 0 and 1).

2. Which of the following experiments have equality 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.

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

Sol. (i) Not equally likely.

(ii) Not equally likely.

(iii) Equality likely, because if a answer us true, then it cannot the take and it answer is false, then it cannot the true. So both have equal possibility either true or false.

(iv) Equality likely, because when a baby is born, they have both equal possibility either born baby is a boy or a girl.

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?

Sol. When we toss a coin, we get either head or tail which are equally likely. Hence, the individual result of a coin toss is completely unpredictable or unbiased.

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

$$\textbf{(a)}\space\frac{\textbf{2}}{\textbf{3}}$$

(b) – 1.5

(c) 15%

(d) 0.7

Sol. (b) – 1.5

Explanation:

The probability of an event cannot be negative at any case.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Sol. Given, P(E) = 0.05

We know that,

$$\text{P(E) +}\space \text{P}( \bar{\text{E}})=1\\\Rarr\space\text{P}( \bar{\text{E}})=1-0.05$$

=  0.95 Play Video about Chapter 15 Probability

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?

Sol. (i) A = an orange flavoured candy

n(A) = 0, n(S) = n(S) (sample space)

$$\text{P(A)}=\frac{n(\text{A})}{n(\text{S})}=\frac{\text{O}}{n(\text{S})}=0$$

(ii) Let A = a lemon flavoured candy

$$n(\text{A})=n(\text{S}),\space n(\text{S})=n(\text{S})\\\text{P(A)}=\frac{n(\text{A})}{n(\text{S})}=\frac{n(\text{S})}{n(\text{S})}=1$$

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?

Sol. Let E = Event of 2 students not having the same birthday

∴ P(E) = 0.992

$$\because\space\text{P(E)+}\text{P(}\bar{\text{E}})=1\\\therefore\space 0.992 +\text{P(}\bar{\text{E}})=1\\\Rarr\space\text{P(}\bar{\text{E}})=1-0.992$$

= 0.008

So, the probability of 2 students having the same birthday is 0.008.

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?

Sol. (i) Let A = The ball drawn is red.

n(A) = 3, n(S) = red balls + Black ball

= 3 + 5 = 8

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}\\=\frac{3}{8}$$

(ii) Let B = Not red ball

n(B) = 5, n(S) = 8

$$\text{P(B)}=\frac{\text{n(B)}}{\text{n(S)}}\\\text{P(B)}=\frac{5}{8}$$

9. A box contains 5 red marbles, 8 white marbels 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?

Sol. n(S) = Total possible outcomes

n(S) = Red marbles + White marbles
+ Green marbles

n(S) = 5 + 8 + 4

n(S) = 17

(i) Let E1 = Taken out red marbles

n(E1) = 5

$$\text{P(E}_1)=\frac{\text{n(E}_1)}{n(\text{S})}\\=\frac{5}{17}$$

(ii) Let E2 = Taken out white marbles.

n(E2) = 8

$$\text{P(E}_2)=\frac{n(\text{E}_2)}{n(\text{S})}\\=\frac{8}{17}$$

(iii) Let E3 = Taken out not green marbles

n(E3) = red marbles + white marbles

= 5 + 8

= 13

$$\text{P(E}_3)=\frac{n(\text{E}_3)}{n(\text{S)}}\\=\frac{13}{17}$$

10. A piggy bank contains hundred 50 paise 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 paise coin? (ii) will not be a ₹ 5 coin?

Sol. Total possible outcomes = Number of 50 p coins + No. of ₹ 1 coins + No. of ₹ 2 coins + No. of ₹ 5 coins

= 100 + 50 + 20 + 10

= 180

P(will fall out 50 paise coin)

$$=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\=\frac{100}{180}\\=\frac{10}{18}\\=\frac{5}{9}$$

(ii) P(will not be a ₹ 5 coin)

$$=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\=\frac{100+50+20}{180}\\=\frac{170}{180}\\=\frac{17}{18}$$

11. Gopi buys a fish from a shop for the aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see figure). What is the probability that the fish taken out is a male fish? Sol. Total number of fish in

an aquarium = 5 male fish + 8 female fish = 13 fish

∴ Probability of taken out a male fish

$$=\frac{\text{Number of male fish}}{\text{Total number of fish}}\\=\frac{5}{13}$$

12. A game of chance consists of spinning at arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7 and 8 (see figure) 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?

Sol. n(S) = Total number of points

n(S) = 8

(i) A1 = The arrow comes at number 8

n(A1) = 1, n(S) = 8

$$\text{P(A}_1) =\frac{\text{n(A}_1)}{n(\text{S})}\\\text{P(A}_1)=\frac{1}{8}$$

(ii) A2 = The arrow comes at an odd number

n(A2) = 1, 3, 5, 7

n(A2) = 4

n(S) = 8

$$\text{P(A}_2)=\frac{\text{n(A}_2)}{\text{n(S)}}\\=\frac{4}{8}\\=\frac{1}{2}$$

(iii) A3 = The arrow comes a number greater than 2

n(A3) = 3, 4, 5, 6, 7, 8

n(A3) = 6

n(S) = 8

$$\text{P(A}_3)=\frac{\text{n(A}_3)}{\text{n(S)}}\\=\frac{6}{8}\\=\frac{3}{4}$$

(iv) A4 = The arrow comes a number less than 9

n(A4) = 1, 2, 3, 4, 5, 6, 7, 8

n(A4) = 8

n(S) = 8

$$\text{P(A}_4) =\frac{\text{n(A}_4)}{n(S)}\\=\frac{8}{8}$$

=   1

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.

Sol. In a die there are six numbers {1, 2, 3, 4, 5, 6}

∴ Total number of possible outcomes, n(S) = 6

(i) Let E1 = Event of getting a prime number = {2, 3, 5}

∴ n(E1) = 3

Probability of getting a prime number

$$\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{3}{6}=\frac{1}{2}$$

(ii) Let E2 = Event of getting a number lying between 2 and 6

= {3, 4, 5}

∴  n(E2) = 3

Probability of getting a number lying between 2 and 6

n(E2) = (3, 4, 5) = 3

$$\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}\\=\frac{3}{6}=\frac{1}{2}$$

(iii) Let E3 = Event of getting an odd number

= {1, 3, 5}

∴ n(E3) = 3

Probability of getting on odd number

$$\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{3}{6}=\frac{1}{2}$$

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

(vi) the queen of diamonds.

Sol. The number of cards in one deck of cards, n(S) = 52

(i) Let E1 = Event of getting a kind of red colour

∴ n(E1) = 2

(∵ In a deck of cards, 26 cards are red and 26 cards are black. There are four kings in a deck in which two are red and two are black)

Probability of getting a king of red colour

$$\text{P(E}_1)=\frac{n(\text{E}_1)}{\text{n(S)}}\\=\frac{2}{52}=\frac{1}{26}$$

(ii) Let E2 Event of getting a face card
∴ n(E2) = 12

[∵ In a deck of cards, there are 12 face card (4 king, 4 jack, 4 queen)]

Probability of getting a face card

$$\text{P(E}_2)=\frac{n(\text{E}_2)}{\text{n(S)}}\\=\frac{12}{52}=\frac{3}{13}$$

(iii) Let E3 = Event of getting a red face card

∴ n(E3) = 6

[∵ In a deck of cards, there are 12 face cards (6 red, 6 black)]

Probability of getting a red face card

$$\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{6}{52}=\frac{3}{26}$$

(iv) Let E4 = Event of getting a jack of hearts

∴ n(E4) = 1

[(∵ There are four jack in a deck (1 heart, 1 club, 1 spade, 1 diamond)]

Probability of getting a jack of heart

$$\text{P(E}_4)=\frac{\text{n(E}_4)}{\text{n(S)}}=\frac{1}{52}$$

(v) Let E5 = Event of getting a spade

∴ n(E5) = 13

(∵ In a deck of cards, 13 spade, 13 club, 13 heart, 13 diamond)

$$\text{P(E}_5)=\frac{\text{n(E}_5)}{\text{n(S)}}\\=\frac{13}{52}=\frac{1}{4}$$

(vi) Let E6 = Event of getting a queen of diamond

∴ n(E6) = 1

(∵ In 13 diamond cards, there is only one queen)

Probability of getting a queen of diamond

$$\text{P(E}_6)=\frac{\text{n(E}_6)}{\text{n(S)}}=\frac{1}{52}$$

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?

Sol. (i) Total number of cards = 5

$$\therefore\space\text{P (picking a queen card) =}\frac{1}{5}$$

(ii) Suppose, queen is drawn and put aside. So, their are four cards left.

(a) P(the second card picked up an ace)

$$=\frac{1}{4}$$

(b) P (the second card picked up an queen)

$$=\frac{0}{4}=0\space$$

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.

Sol. Let A = Event of selecting a good pen

n(S) = Total number of pens

n(S) = 12 defective + 132 good pen

n(S) = 144

n(A) = 132

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}\\=\frac{132}{144}\\=\frac{11}{12}$$

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?

Sol. (i) Total number of bulbs, n(S) = 20

Let E1 = Event of getting a defective bulb

n(E1) = 4

∴ Probability of getting a defective bulb

$$=\frac{n(\text{E}_1)}{\text{n(S)}}=\frac{4}{20}=\frac{1}{5}$$

(ii) Suppose, one good bulb is drawn and put outside

Therefore, 15 good bulbs and 4 defective bulbs are remain in a lot.

Now, total number of bulbs n(S1) = 19

Let E2 = Event of selecting a good bulb,

n(E2) = 15

$$\therefore\text{Required probability =}\frac{\text{n(E}_2)}{\text{n(S}_1)}=\frac{15}{19}$$

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

Sol. (i) Total number of disc in a box n(S) = 90

Let E1 = Event of selecting a two digit number

n(E1) = {10, 11, ...., 90}

∴ n(E1) = 81

Probability of getting a two digit number

$$\text{P(E}_1)=\frac{\text{n(E)}}{\text{n(S)}}\\=\frac{81}{90}=\frac{9}{10}$$

(ii) Let E2 = Event of selecting a perfect square number

n(E2) = {1, 4, 9, 16, 25, 36, 49, 64, 81}

∴ n(E2) = 9

Probability of selecting a perfect square number

$$\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}\\=\frac{9}{90}=\frac{1}{10}$$

(iii) Let E3 = Event of selecting a number divisible by 5

= {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 80}

∴ n(E3) = 18

∴ Required probability

$$\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{18}{90}=\frac{1}{5}$$

19. A child has a die whose six faces show the letters as given below. The die is thrown once. What is the probability of getting (i) A? (ii) D?

Sol. Total number of outcomes of a six face die, n(S) = 6

(i) Let A1 = Event of getting a letter A

∴ n(A1) = 2

Probability of getting a letter A

$$\text{P(A}_1)=\frac{\text{n(A}_1)}{\text{n(S)}}\\=\frac{2}{6}=\frac{1}{3}$$

(ii) Let A2 = Event of getting a letter D

∴ n(A2) = 1

Probability of getting a letter D

$$\text{P(A}_2)=\frac{\text{n(A}_2)}{\text{n(S)}}=\frac{1}{6}$$

20. Suppose, you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m? Sol. Total number of cases, n(s) = Area of rectangle

= 3 × 2 = 6m2

Favourable number of cases, n(E) = Area of circle

$$=\pi\bigg(\frac{1}{2}\bigg)^{2}=\frac{\pi}{4}m^{2}\\\bigg(\because\space\text{Diameter = 1}\Rarr\text{radius}=\frac{1}{2}\bigg)$$

∴ Probability that the die land inside the circle

$$\text{P(E)}=\frac{n(\text{E})}{n(\text{S})}\\=\frac{\pi/4}{6}=\frac{\pi}{24}$$

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

(ii) she will not buy it?

Sol. Total number of ball pens, n(S) = 144

(i) Let E = Event of getting a good ball pen

∴ n(E) = 144 – 20 = 124

She will buy it only if it is a good pen.

$$\therefore\space\text{Required probability =}\frac{\text{n(E)}}{\text{n(S)}}\\=\frac{124}{144}=\frac{31}{36}$$

(ii) She will not buy if it not a good pen.

∴ Required probability = P(not E)

P(E) + P(not E) = 1

$$\text{P(not E) =}\space1-\frac{31}{36}=\frac{5}{36}$$

22. Two dice, one blue and one grey are thrown at the same time

(i) complete the following table.

 Event sum of 2 dice 2 3 4 5 6 7 8 9 10 11 12 Probability $$\frac{1}{36}$$ $$\frac{5}{36}$$

(ii) A student argues that these are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of the them has a probability $$\frac{1}{11}$$. Do you agree with this argument ? Justify your answer.
Sol. (i)
Total possible outcomes of throwing the two dice, S

= [(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

∴ n(S) = 36

(a) Let A1 = Sum of two dice is 3 = (1, 2), (2, 1)

n(A1) = 2

$$\therefore\space \text{P(A}_1)=\frac{\text{n(A}_1)}{\text{n(S)}}\\=\frac{2}{36}=\frac{1}{18}$$

(b) Let A2 = Sum of two dice is 4 = {(1, 3), (2, 2), (3, 1)

n(A2) = 3

$$\therefore\space\text{P(A}_2)=\frac{\text{n(A}_2)}{\text{n(S)}}\\=\frac{3}{36}=\frac{1}{12}$$

(c) Let A3 = Sum of two dice is 5 = {(1, 4), (2, 3), (3, 2), (4, 1)}

n(A3) = 4

$$\therefore\space\text{P(A}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{4}{36}=\frac{1}{9}$$

(d) Let A4 = Sum of two dice is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)]

n(A4) = 5

$$\therefore\space\text{P(A}_4)=\frac{5}{36}$$

(e) Let A5 = Sum of two dice is 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)]

n(A5) = 6

$$\therefore\space\text{P(A}_5)=\frac{\text{n(A}_5)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(f) Let A6 = Sum of two dice is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)]

n(A6) = 5

$$\therefore\space\text{P(A}_6)=\frac{\text{n(A}_6)}{\text{n(S)}}=\frac{5}{36}$$

(g) Let A7 = Sum of two dice is 9 = {(3, 6), (4, 5), (5, 4), (6, 3)]

n(A7) = 4

$$\therefore\space\text{P(A}_7)=\frac{\text{n(A}_7)}{\text{n(S)}}\\=\frac{4}{36}=\frac{1}{9}$$

(h) Let A8 = Sum of two dice is 10 = {(4, 6), (5, 5), (6, 4)]

n(A8) = 3

$$\therefore\space\text{P(A}_8)=\frac{\text{n(A}_8)}{\text{n(S)}}\\=\frac{3}{36}=\frac{1}{12}$$

(i) Let A9 = Sum of two dice is 11 = [(6, 5), (5, 6)]

n(A9) = 2

$$\therefore\space\text{P(A}_9)=\frac{\text{n(A}_9)}{\text{n(S)}}\\=\frac{2}{36}=\frac{1}{18}$$

(j) Let A10 = Sum of Two dice is 12 = {(6, 6)}

n(A10) = 1

$$\therefore\space\text{P(A}_{10})=\frac{\text{n(A}_{10})}{\text{n(S)}}=\frac{1}{36}$$

(ii) No, we do not agree with the argument because these events are not equally likely.

23. A game consists of tossing a one rupee coin 3 times and nothing its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads of three tails and losses otherwise. Calculate the probability that Hanif will lose the game.

Sol. n(S) = Total possible outcomes

n(S) = [(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)]

A = Hanif wins the game

Not A = Hanif will loses the game

n(A) = [(H, H, H), (T, T, T)]

n(A) = 2

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}\\=\frac{2}{8}=\frac{1}{4}$$

We know that

P(A) + P(not A) = 1

Then

$$\frac{1}{4}+\text{P(notA) = 1}\\\Rarr\space\text{P(not A) =}1-\frac{1}{4}\\\Rarr\space\text{P(not A) =}\frac{3}{4}\\\therefore\space\text{Required probability =}\frac{3}{4}.$$

24. A die is thrown twice. What is the probability that

(i) 5 will not come up either time ?

(ii) 5 will come up atleast once?

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

Sol. (i) Total number of cases, n(S) = 62 = 36

$$\text{Let}\space\bar{\text{E}}=\text{Event that 5 will come up either time}$$

= [(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),
(5, 2), (5, 3), (5, 4), (5, 6)

$$\Rarr\space\text{n}(\bar{\text{E}})=11$$

and E = Event that 5 will not come up either time

∴ n(E) = 36 – 11 = 25

∴ Probability that 5 will not come up either

$$\text{time}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{25}{35}$$

(ii) Probability that, 5 will come up alteast once

$$\text{P}(\bar{\text{E}})=\frac{\text{n}(\bar{\text{E}})}{\text{n(S)}}=\frac{11}{35}$$

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 $$\frac{1}{3}.$$

(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 $$\frac{1}{2}.$$

Sol. (i) Incorrect.

The possible outcomes of tossing two coins

= {(H, H), (H, T), (T, H), (T, T)}

$$\text{Then, P(H, H) =}\frac{1}{4},\text{P(T,T)}=\frac{1}{4}\\\text{and P{(H, T), (T, H)} =}\frac{2}{4}=\frac{1}{2}$$

(ii) Correct. The total possible outcomes on a die = {1, 2, 3, 4, 5, 6}

Out of them odd numbers are {1, 3, 5} and even numbrs are {2, 4, 6}

$$\therefore\space\text{P(getting odd numbers) =}\frac{3}{6}=\frac{1}{2}\\\text{and P(getting even numbers) =}\frac{3}{6}=\frac{1}{2}\\\text{Hence, both have same probability}\space\frac{1}{2}.$$

Exercise 15.2 (Optional)

1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) difference days?

Sol. Number of days from Tuesday to Saturday = 5

Total number of possible outcomes = 5 × 5 = 25

(i) If both visit the shop on the same day, then number of favourate cases.

n(E) = 5

∴ P(both will visit on the same day)

$$=\frac{5}{25}=\frac{1}{5}$$

(ii) If they visits the shop on consecutive days, then number of cases

= 8 (T, W : W, Th : Th, F : F, S : S, F : F, Th : Th, W : W, T)

∴ P(both will visit on consecutive days)

$$=\frac{8}{25}$$

(iii) P(both will visit on different days)

= 1 – P(both will visit on the same day)

$$=\space 1-\frac{1}{5}=\frac{4}{5}$$

2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws : What is the probability that the total score is

(i) even? (ii) 6? (iii) atleast 6?

Sol. The complete table is Total number of possible outcomes = 36

(i) Let E1 = event that total score is even

∴   n(E1) = 18

$$\text{P(total score in even) =}\frac{18}{36}=\frac{1}{2}$$

(ii) Let E2 = event that total score is 6.

∴ n(E2) = 4

$$\text{P(total score is 6) =}\frac{4}{36}=\frac{1}{9}$$

(iii) Let E3 = event that total score is at least 6, i.e., 6, 7, 8, 9, 12

∴ n(E2) = 15

$$\text{P(total score is at least 6) =}\frac{5}{36}=\frac{5}{12}$$

3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Sol. Let the number of blue balls = x

∴ Total number of balls = 5 + x

A = drawing a red ball

n(A) = 5

n(S) = 5 + x

$$\text{P(A)}=\frac{5}{5+x}$$

B = drawing a blue balls

n(B) = x

$$\text{P(B)}=\frac{\text{n(B)}}{\text{n(S)}}=\frac{x}{5+x}$$

According to question

P(B) = 2P(A)

$$\frac{x}{5+x}=2×\frac{5}{5+x}\\\Rarr\space\frac{x}{5+x}=\frac{10}{5+x}$$

⇒ x = 10

Hence, the number of blue balls = 10.

4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Sol. Given, number of black balls = x

and total number of balls = 12

∴ Probability of drawing a ball which is black

$$\text{P}_1=\frac{\text{Number of black balls}}{\text{Total number of balls}}=\frac{x}{12}$$

If 6 more black balls are put in the box, then total number of black balls = x + 5 and total number of balls = 12 + 6 = 18

∴ Probability of drawing a black ball,

$$\text{P}_2=\frac{x+16}{18}$$

Now, according to the question,

P2 = 2P1

$$\Rarr\space\frac{x+6}{18}=2×\frac{x}{12}\\\Rarr\space\frac{x+6}{18}=\frac{x}{6}$$

⇒ 6(x + 6) = 18x

⇒ x + 6 = 3x

⇒ 2x = 6

⇒ x = 3

5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $$\frac{2}{3}.$$ Find the number of blue balls in the jar.

Sol. Let number of green marbles = x

Then, number of blue marbles = 24 – x

Total number of marbles = 24

Now, probability of drawing a green marble

$$=\frac{\text{Number of green marbles}}{\text{Total number of marbles}}\\=\frac{2}{3}\space\text{(Given)}\\\Rarr\space\frac{x}{24}=\frac{2}{3}$$

⇒ 3x = 24 × 2

$$\Rarr\space x=\frac{24×2}{3}$$

⇒ x = 8 × 2

x = 16

∴ Number of green marbles = 16

Hence, number of blue marbles = 24 – 16 = 8

Selected NCERT Exemplar Problems

Exercise 15.1

• Choose the correct answer from the given four options.

1. If an event cannot occur, then its probability is

(a) 1

$$\textbf{(b)\space}\frac{\textbf{3}}{\textbf{4}}$$

$$\textbf{(c)\space}\frac{\textbf{1}}{\textbf{2}}$$

(d) 0

Sol. (d) 0

Explanation:

If an event cannot occur, then it is impossible event and probability of impossible event is zero.

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

$$\textbf{(a)}\space\frac{\textbf{1}}{\textbf{3}}$$

(b) 0.1

(c) 3

$$\textbf{(d)}\space\frac{\textbf{17}}{\textbf{16}}\\\textbf{Sol.}\space\text{(d)}\space\frac{17}{16}$$

Explanation:

Since, probability of an event always lie between 0 and 1.

3. An event is very unlikely to happen. Its probability is closest to

(a) 0.0001

(b) 0.001

(c) 0.01

(d) 0.1

Sol. (a) 0.0001

Explanation:

The probability of an event which is very unlikely to happen is closest to zero from the given options 0.0001 is closest to zero.

4. If the probability of an event is P, the probability of its complementary event will be

(a) P – 1

(b) P

(c) 1 – P

$$\textbf{(d)\space}\textbf{1}-\frac{\textbf{1}}{\textbf{P}}$$

Explanation:

Since, probability of an event + Probability of its complementary event = 1

∴ Probability of its complementary event

= 1 – Probability of an event = 1 – P

5. The probability expressed as a percentage of a particular occurrence can never be

(a) less than 100

(b) less than 0

(c) greater than 1

(d) anything but a whole number

Sol. (b) less than 0

Explanation:

We know that the probability expressed as a percentage always lie between 0 and 100. So, it cannot be less than 0.

6. If P(A) denote the probability of an event A, then

(a) P(A) < 0

(b) P(A) > 1

(c) 0 ≤ P(A) ≤ 1

(d) – 1 ≤ P(A) ≤ 1

Sol. (c) 0 ≤ P(A) ≤ 1

Explanation:

Since, probability of an event always lie between 0 and 1.

7. A card is selected from a deck of 52 cards. The probability of its being a red face card is

$$\textbf{(a)}\space\frac{3}{26}$$

$$\textbf{(b)}\space\frac{\textbf{3}}{\textbf{13}}$$

$$\textbf{(c)}\space\frac{\textbf{2}}{\textbf{13}}$$

$$\textbf{(d)}\space\frac{\textbf{1}}{\textbf{2}}$$

$$\textbf{Sol.}\space\text{(a)}\space\frac{3}{26}$$

Explanation:

In a deck of 52 cards, there are 12 face cards 6 red and 6 black

∴ Probability of getting a red face card

$$=\frac{6}{52}=\frac{3}{26}$$

8. The probability that a non leap your selected at random will contains 53 sundays is

$$\textbf{(a)}\space\frac{\textbf{1}}{\textbf{7}}$$

$$\textbf{(b)\space}\frac{\textbf{2}}{\textbf{7}}$$

$$\textbf{(c)\space}\frac{\textbf{3}}{\textbf{7}}$$

$$\textbf{(d)\space}\frac{\textbf{5}}{\textbf{7}}$$

$$\textbf{Sol.\space}(a)\space\frac{1}{7}$$

Explanation:

A non leap year has 365 days and therefore 52 weeks and 1 day. This 1 day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.

Thus, out of 7 possibilities, 1 favour the event that the one day is Sunday.

$$\therefore\space\text{Required probability =}\frac{1}{7}$$

9. When a die is thrown, the probability of getting an odd number less than 3 is

$$\textbf{(a)}\space\frac{\textbf{1}}{\textbf{6}}$$

$$\textbf{(b)}\space\frac{\textbf{1}}{\textbf{3}}$$

$$\textbf{(c)}\space\frac{\textbf{1}}{\textbf{2}}$$

$$\textbf{(d)}\space\textbf{0}$$

$$\textbf{Sol.}\space\text{(a)}\space\frac{1}{6}$$

Explanation:

When a die is thrown then total number of outcomes = 6

Odd number less than 3 is 1 only.

∴ Number of possible outcomes = 1

$$\text{Required probability =}\frac{1}{6}$$

10. A card is drawn from a deck of 52 cards, the event E in that card is not an ace of hearts. The number of outcomes favourable to E is

(a) 4

(b) 13

(c) 48

(d) 51

Sol. (d) 51

Explanation:

In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.

∴ The number of outcomes favourable to E = 51

11. The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is

(a) 7

(b) 14

(c) 21

(d) 28

Sol. (b) 14

Explanation:

Here, total number of eggs = 400

Probability of getting a bad egg = 0.035

$$\Rarr\space\frac{\text{Number of bad eggs}}{\text{Total number of eggs}}=0.035\\\Rarr\space\frac{\text{Number of bad eggs}}{400}=0.035$$

⇒ Number of bad eggs = 0.035 × 400 = 14

12. A girl calculates that the probability of her wining the first prize in a lottery is 0.08. If 600 tickets are sold, how many tickets has she bought?

(a) 40

(b) 240

(c) 480

(d) 750

Sol. (c) 480

Explanation:

Given, total number of sold tickets = 6000

Let she brought x tickets.

Then probability of her winning the first prize

$$=\frac{x}{6000}=0.08\space\text{(Given)}$$

⇒ x = 0.08 × 6000

⇒ x = 480

Hence, she bought 480 tickets.

13. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

$$\textbf{(a)\space}\frac{\textbf{1}}{\textbf{5}}$$

$$\textbf{(b)\space}\frac{\textbf{3}}{\textbf{5}}$$

$$\textbf{(c)\space}\frac{\textbf{4}}{\textbf{5}}$$

$$\textbf{(d)\space}\frac{\textbf{1}}{\textbf{3}}$$

$$\textbf{Sol.\space}\text{(a)}\space\frac{1}{5}$$

Explanation:

Number of total outcomes = 40

Multiply of 5 between 1 to 40 = 5, 10, 15, 20, 25, 30, 35, 40

∴ Total number of possible outcomes = 6

$$\text{Required probability =}\frac{8}{40}=\frac{1}{5}$$

14. Someone is asked to take a number from 1 to 100. The probability that it is a prime, is

$$\textbf{(a)}\space\frac{\textbf{1}}{\textbf{5}}$$

$$\textbf{(b)}\space\frac{\textbf{6}}{\textbf{25}}$$

$$\textbf{(c)}\space\frac{\textbf{1}}{\textbf{4}}$$

$$\textbf{(d)}\space\frac{\textbf{13}}{\textbf{50}}$$

$$\textbf{Sol.}\space\text{(c)}\space\frac{1}{4}$$

Explanation:

Total number of outcomes = 100

Prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61,67, 71, 73, 79, 83, 89 and 97.

∴ Total number of possible outcomes = 25

$$\text{Required probability =}\frac{25}{100}=\frac{1}{4}$$

15. A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is

$$\textbf{(a)\space}\frac{\textbf{4}}{\textbf{23}}$$

$$\textbf{(b)\space}\frac{\textbf{6}}{\textbf{23}}$$

$$\textbf{(c)\space}\frac{\textbf{8}}{\textbf{23}}$$

$$\textbf{(d)\space}\frac{\textbf{17}}{\textbf{23}}$$

$$\textbf{Sol.}\space\text{(b)}\space\frac{6}{23}$$

Explanation:

Total number of students = 23

Number of students in house A, B and C = 4
+ 8 + 5 = 17

Remains students = 23 – 17 = 6

∴ Probability that the selected student is not from A, B and C

$$=\frac{6}{23}.$$

Exercise 15.2

1. In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $$=\frac{1}{4}.$$ Is this correct? Justify your answer.

Sol. No. The probability of each is not $$=\frac{1}{4}.$$ because the probability of no girl in three children is zero and probability of three girls in three children is one.

Justification: So these events are not equally likely as outcome one girl, means gbb, hbh, bbh is outcome one girl is possible in only one way whereas outcome one boy is possible in two ways.

2. A game consists of spinning an arrow which comes to rest pointing at one of the regions

(1, 2 or 3) (see figure). Are the outcomes 1, 2, and 3 equally likely to occur? Give reasons. Sol. No. The outcomes are not equally likely, because ‘3’ contains half part of the total region so it is more likely than 1 and 2 since 1 and 2 contains half part of the remain part of the region.

3. Apooru throws two dice once and completes 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?

Sol. Apooru throws two dice once.

So, total number of outcomes = 36

Number of outcomes for getting product 36 = 1(6 × 6)

$$\therefore\space\text{Probability for Apporu =}\frac{1}{36}$$

Also, Peehu throws one die,

So total number of outcomes = 6

Number of outcomes for getting square 36 = 1(62 = 36)

$$\therefore\space\text{Probability for Peehu =}\frac{1}{6}=\frac{6}{36}$$

Hence Peehu has better chance to getting the number 36.

4. When we toss a coin, there are two possible oucomes—Head or Tail. Therefore, the probability of each outcome is $$\frac{\textbf{1}}{\textbf{2}}.$$ Justify your answer.

Sol. Yes. Probability of each outcomes $$=\frac{1}{2}$$ because head and tail both are equally likely events.

5. 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 $$=\frac{1}{2}$$. Is this correct? Give reasons.

Sol. No. This is not correct.

If we throw a die then total number of outcomes = 6

Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6

$$\therefore\space\text{Probability of getting 1 =}\frac{1}{6}$$

Now, probability of getting ‘not 1’

= 1 – Probability of getting 1

$$=1-\frac{1}{6}=\frac{1}{5}$$

6. I toss three coins together. The possible outcomes are no heads, 1 head, 2 head and 3 head. So, I saythat probability of no heads is $$\frac{\textbf{1}}{\textbf{4}}.$$ What is wrong with this conclusion?

Sol. I toss three coins together

So, total number of outcomes = 23 = 8

and possible outcomes are (HHH), (HTT), (THT), (TTH), (HHT), (THH), (HTH) and (TTT)

Now, probability of getting no head

$$=\frac{1}{6}$$

Hence, the given conclusion is wrong because the probability of no head is

$$\frac{1}{8}\space\text{not the}\space\frac{1}{4}.$$

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

Sol. No. If we toss a coin then we get head or tail, both are equally likely events. So, probability is $$\frac{1}{2}.$$

If we toss a coin 6 times then probability will same in each case. So, the probability of getting a head is not 1.

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

Sol. The outcome of next toss may or may not be tail, because on tossing a coin we get head or tail so both are equally likely events.

9. A bay 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 $$\frac{1}{2}.$$ Justify.

Sol. We know that between 1 to 100 half numbers are even and half numbers are odd i.e., 50 numbers (2, 4, 6, 8, ...... 96, 98, 100) are even and 50 numbers (1, 3, 5, 7, .... 97, 99) are odd. So, both events are equally likely.

So, probability of getting even numbers

$$=\frac{50}{100}=\frac{1}{2}$$

and probability of getting odd numbers

$$=\frac{50}{100}=\frac{1}{2}$$

Hence, probability of each is $$\frac{1}{2}.$$

Exercise 15.3

1. Two dice are thrown at the same time. Find the probability of getting

(i) same number on both dice.

(ii) difference number on both dice.

Sol. Two dice are thrown at the same time.

So, total number of outcomes = 36

(i) We get same number on both dice.

So, possible outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)

∴ Number of possible outcomes = 6

$$\text{Now, required probability =\space}\frac{6}{36}=\frac{1}{6}$$

(ii) We get different number on both sides.

So, number of possible outcomes

= 36 – number of possible outcomes for same number on both dice

= 36 – 6 = 30

$$\therefore\space\text{Required probability =}\frac{30}{36}=\frac{5}{6}$$

2. 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?

Sol. Two dice are thrown simultaneously

So, number of possible outcomes = 36

(i) Sum of the numbers appearing on the dice is 7.

Possible ways are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

Number of possible ways = 6

$$\therefore\space\text{Required probability =}\frac{6}{36}=\frac{1}{6}$$

(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

∴ Possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).

Number of possible ways = 15

$$\therefore\space\text{Required probability =}\frac{15}{36}=\frac{5}{12}$$

(iii) Sum of the numbers appearing on the dice is 1.

It is not possible, so its probability is zero.

3. Two dice are thrown together. Find the probability that the product of the numbers ont he top of the dice is

(i) 6

(ii) 12

(iii) 7

Sol. Number of total outcomes = 36

(i) When product of the numbers on the top of the dice is 6.

Possible ways are (1, 6), (2, 3), (3, 2), (6, 1)

$$\therefore\space\text{Required probability =}\frac{4}{36}=\frac{1}{9}$$

(ii) When product of the numbers on the top of the dice is 12.

Possible ways are (2, 6), (3, 4), (4, 3), (6, 2).

Number of possible ways = 4

$$\therefore\space\text{Required probability =}\frac{4}{36}=\frac{1}{9}$$

(iii) Product of the numbers on the top of the dice cannot be 7. So, is probability is zero.

4. 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.

Sol. Number of total outcomes = 36

When product of numbers appearing on them is less than 9, then possible ways are (1, 6), (1, 5), (1, 4), (1, 3), (1, 2), (1, 1), (2, 2), (2, 3), (2, 4), (3, 2), (4, 2), (4, 1), (3, 1), (5, 1), (6, 1) and (2, 1)

Number of possible ways = 16

$$\therefore\space\text{Required probability =}\frac{16}{36}=\frac{4}{9}$$

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

Sol. Number of total outcomes = 62 = 36

(i) Let E1 = Event of getting sum 2 = {(1, 1), (1, 1)}

n(E1) = 2

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{2}{36}=\frac{1}{18}$$

(ii) E2 = Event of getting sum 3 = {(1, 2), (1, 2), (2, 1), (2, 1)}

n(E2) = 4

$$\therefore\space\text{P(E}_2)=\frac{n(\text{E}_2)}{\text{n(S)}}\\=\frac{4}{36}=\frac{1}{9}$$

(iii) Let E3 = Event of getting sum 4 = {(2, 2), (2, 2), (3, 1), (3, 1), (1, 3), (1, 3)}

∴ n(E3) = 6

$$\therefore\space\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(iv) Let E4 = Event of getting sum 5 = {(2, 3), (2, 3), (4, 1), (4, 1), (3, 2), (3, 2)}

∴ n(E4) = 6

$$\therefore\space\text{P(E}_4)=\frac{\text{n(E}_4)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(v) Let E5 = Event of getting sum 6 = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}

n(E5) = 6

$$\therefore\space\text{P(E}_5)=\frac{\text{n(E}_5)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(vi) Let E6 = Event of getting sum 7 = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}

∴ n(E6) = 6

$$\therefore\space\text{P(E}_6)=\frac{\text{n(E}_6)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(vii) Let E7 = Event of getting sum 8 = {(5, 3), (5, 3), (5, 2), (6, 2)}

∴ n(E1) = 4

$$\therefore\space\text{P(E}_7)=\frac{\text{n(E}_7)}{\text{n(S)}}\\=\frac{4}{36}=\frac{1}{9}$$

(viii) Let E8 = Event of getting sum 9 = {(6, 3), (6, 3)}

∴ n(E8) = 2

$$\therefore\space\text{P(E}_8)=\frac{\text{n(E}_8)}{\text{n(S)}}\\=\frac{2}{36}=\frac{1}{18}$$

6. A coin is tossed 3 times. List the possible outcomes. Find the probability of getting

Sol. The possible outcomes of a coin is tossed 3 times.

S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}

∴ n(S) = 8

(i) Let E1 =Event of getting all heads = {(HHH)}

∴ n(E1) = 1

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{1}{8}$$

(ii) Let E2 = Event of getting at least 2 heads = {(HHT), (HTH), (THH), (HHH)}

∴ n(E2) = 4

$$\therefore\space\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}\\=\frac{4}{8}=\frac{1}{2}$$

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

Sol. The total number of sample space in two dice, n(S) = 6 × 6 = 36.

Let E = Event of getting the difference of the two numbers is 2.

= {(1, 3), (2, 4),(3, 5), (4, 6), (3, 1), (4, 2),(5, 3), (6, 4)}

∴ n(E) = 8

$$\therefore\space\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}\\=\frac{8}{36}=\frac{2}{9}$$

8. 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

Sol. If we remove one king, one queen and one jack of clubs from 52 cards, then the remaining cards left, n(S) = 49.

(i) Let E1 = Event of getting a heart

n(E1) = 13

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{13}{49}$$

(ii) Let E2 = Event of getting a king

n(E2) = 3

(∵ In out of 4 king, one club cards is already remove)

$$\therefore\space\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{3}{49}$$

9. 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

Sol. In out of 52 play cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining cards are left, n(s) = 52 – 3 × 4 = 40.

(i) Let E1 = Event of getting card value is 7.

E = Card value 7 has either a spade is diamond, a club or a heart

∴ n(E1) = 4

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{4}{40}=\frac{1}{10}$$

(ii) Let E2 = Event of getting card value is 7.

E = Event of getting card value is 8, 9 and 10

∴ n(E2) = 3 × 4

(∵ Each card value has four cards)

= 12

$$\therefore\space\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}\\=\frac{12}{40}=\frac{3}{10}$$

(iii) Let E3 = Event of getting card value is less than 7

= Event of getting card value is 1, 2, 3, 4, 5, 6

∴ n(E3) = 6 × 4

(∵ Each card value has four cards)

= 24

$$\therefore\space\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{24}{40}=\frac{3}{5}$$

10. An integer is chosen between 0 and 100. What is the probability that it is

(i) divisible by 7? (ii) not divisible by 7?

Sol. The number of integers between 0 and 100 is

n(S) = 99

(i) Let E1 = Event of choosing a integer which is divisible by 7
= Event of choosing a integer which is multiply of 7

= {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}

∴ n(E1) = 14

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{14}{99}$$

(ii) E2 = Event of choosing a integer which is not divisible by 7.

∴ n(E2) = n(S) – n(S1)

= 99 – 14 = 85

$$\therefore\space\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{85}{99}$$

11. 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?

Sol. The total number of sealed envelopes in a box is n(S) = 1000

The number of prize envelopes = 10 + 100
+ 200 = 310

The number of unprize envelopes

n(E) = 1000 – 310 = 690

$$\therefore\space\text{P(E) =}\frac{\text{n(E)}}{\text{n(S)}}\\=\frac{690}{1000}=\frac{69}{100}=0.69$$

12. 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 resuffled. A slip is drawn at random. What is the probability that it is marked other than  1?

Sol. Total number of slips in a box, n(S) = 25 + 50 = 75 From the chart it is clean that, 11 slips which are not in ₹ 1.

∴ Required probability

$$=\frac{\text{Number of slips other than ₹ 1}}{\text{Total number of slips}}\\=\frac{11}{75}$$

13. 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?

Sol. Since, total number of bulbs are, n(S) = 24. Let E1 = Event of selecting not defective of bulb = Event of selecting good bulbs.

n(E1) = 18

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{18}{24}=\frac{3}{4}$$

Suppose, the selecting bulb is defective and not replaced, then total bulbs remains in a carton, n(S) = 23.

In those of them 18 good bulbs and 5 defective bulbs.

∴ P(selecting second defective bulb)

$$=\frac{5}{23}.$$

14. 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

Sol. Total number of figures

n(S) = 8 triangles + 10 squares = 18 (i) P(lost piece is a triangle)

$$=\frac{8}{18}=\frac{4}{9}$$

(ii) P(lost piece is a square)

$$=\frac{10}{18}=\frac{5}{9}$$

(iii) P(square of blue colour)

$$=\frac{6}{18}=\frac{1}{3}$$

(iv) P(triangle of red colour)

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

15. In a game, the entry free of ` 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.

Sol. Total number of cases of tossing a coin 3 times

S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)]

∴ n(S) = 8

(i) Let E1 = Event that Sweta losses the entry fee

= she tosses tail on three times

n(E1) = {(TTT)}

$$\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}=\frac{1}{8}$$

(ii) Let E2 = Event that Sweta get double entry fee

= she tosses heads on three times = {(HHH)}

n(E2) = 1

$$\therefore\space\text{P(E}_2) =\frac{\text{n(E}_2)}{\text{n(S)}}=\frac{1}{8}$$

(iii) Let E3 = Event that Sweta gets her entry fee back

= Sweta gets heads one or two times

= {(HTT), (THT), (TTH), (HHT), (HTH), (THH)}

∴ n(E1) = 6

$$\therefore\space\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{6}{8}=\frac{3}{4}$$

16. A die has its six faces marked 0, 1, 1, 1, 5, 6. Two such dice are thrown together and the total score is recorded.

(i) How many different scores are possibles?

(ii) What is the probability of getting a total of 7?

Sol. Given a die has six faces marked {0, 1, 1, 1, 6, 6}

∴ Total sample space, n(S) = 62 = 36

(i) The different score are possible (0, 1, 2, 6, 7)

(ii) Let E = Event of getting a sum 7

= {(1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1)

∴ n(E) = 12

$$\therefore\space\text{P(E) =}\frac{\text{n(E)}}{\text{n(S)}}\\=\frac{12}{36}=\frac{1}{3}$$

17. 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 are random from the lot. What is the probability that it is

(i) acceptable to Varnika?

Sol. Given, total number of mobile phones

(i) Let E1 = Event that varnika will buy a mobile phone

= Varnika buy only, if it is good mobile

∴ n(E1) = 42

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{42}{48}=\frac{7}{8}$$

(ii) Let E2 = Event that trader will buy only when it has no major defects

$$\therefore\space\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}\\=\frac{45}{48}=\frac{15}{16}$$