Jharkhand Board JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 15 Probability Exercise 15.1

Question 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 ………………

Solution:

i) 1,

ii) 0, impossible event,

iii) 1, sure or certain,

iv) 1,

v) 0, 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 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:

(iii) and (iv) have equally likely outcomes. Only two possibilities are there in each of these cases.

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:

When a coin is tossed, head or tail are equally likely possible events. So the result of an individual coin toss is unpredictable.

Question 4.

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

A) \(\frac{2}{3}\)

B) -1.5

C) 15%

D) 0.7?

Solution:

(B) Because, probability of an event cannot be negative.

Question 5.

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

Solution:

P(E) = 0.05.

[P(\(\overline{\mathrm{E}}\)) = Probability of not an event]

We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1

P(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) P(orange flavoured candy) = 0. Impossible event.

ii) P(Lemon flavoured candy) = 1. Sure event.

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:

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

∴ P(E) = 0.992

∴ P(E) + P(\(\overline{\mathrm{E}}\)) = 1

∴ 0.992 + P(\(\overline{\mathrm{E}}\)) = 1

⇒ P(\(\overline{\mathrm{E}}\)) = 1 – 0.992

= 0.008

So, the probability of two students having the same birthday is 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:

Total number of balls, n(S) = 3 + 5 = 8.

Let E = Event of drawing 1 red ball

∴ n(E) = 3

(i) Probability of drawing a red ball = \(\frac{n(E)}{n(S)}=\frac{3}{8}\)

(ii) Probability of not drawing a red ball = 1 – P(Drawing a red ball)

= \(1-\frac{3}{8}=\frac{5}{8}\)

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:

Total possible outcomes = 5 + 8 + 4 = 17.

P(R) = \(\frac{5}{17}\), P(W) = \(\frac{8}{17}\)

P(Not green) = P(R + W) = \(\frac{5}{17}+\frac{8}{17}=\frac{13}{17}\)

Question 10.

A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 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 Rs. 5 coin?

Solution:

Total possible outcomes: 100 + 50 + 20 + 10 = 180.

P(50 paise coin) = \(\frac{100}{180}=\frac{5}{9}\)

P(Not Rs. 5 coin) = \(\frac{100}{180}+\frac{50}{180}+\frac{20}{180}\)

= \(\frac{170}{180}=\frac{17}{18}\).

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. What is the probability that the fish taken out is a male fish?

Solution:

Total number of fish in an aquarium = 5 male fish + 8 female fish = 13 fish

∴ Probability of taking out a male fish = \(\frac{\text { Number of male fish }}{\text { Total number of fish }}=\frac{5}{13}\)

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, 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:

Total possible outcomes = 8.

i) P(8) = \(\frac{1}{8}\)

ii) P(odd number) = \(\frac{4}{8}=\frac{1}{2}\)

iii) P(no. > 2) = \(\frac{6}{8}=\frac{3}{4}\)

iv) P(no. < 9) = \(\frac{8}{8}\) = 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:

Total possible outcomes 1, 2, 3, 4, 5, 6 = 6

P(Prime number) (2, 3, 5) = \(\frac{3}{6}=\frac{1}{2}\)

P(Number between 2 and 6) = \(\frac{3}{6}=\frac{1}{2}\)

P(Odd number) = \(\frac{3}{6}=\frac{1}{2}\)

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:

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

i) Let E_{1} = Event of getting a king of red colour

∴ n(E_{1}) = 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 = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{52}=\frac{1}{26}\)

ii) Let E_{2} = Event of getting a face card

∴ n(E_{2}) = 12

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

Probability of getting a face card = \(\frac{n\left(E_2\right)}{n(S)}=\frac{12}{52}=\frac{3}{13}\)

iii) Let E_{3} = Event of getting a red face card

∴ n(E_{3}) = 6

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

Probability of getting a red face card = \(\frac{n\left(E_3\right)}{n(S)}=\frac{6}{52}=\frac{3}{26}\)

iv) Let E_{4} = Event of getting a jack of hearts

∴ n(E_{4}) = 1

(∵ There are four jacks in a deck- 1 heart, 1 club, 1 spade, 1 diamond)

Probability of getting a jack of hearts = \(\frac{n\left(E_4\right)}{n(S)}=\frac{1}{52}\)

v) Let E_{5} = Event of getting a spade

∴ n(E_{5}) = 13

(∵ In a deck of cards, there are 13 spades, 13 clubs, 13 hearts, 13 diamonds)

Probability of getting a spade = \(\frac{n\left(E_5\right)}{n(S)}=\frac{13}{52}=\frac{1}{4}\)

vi) Let E_{6} = Event of getting a queen of diamond

∴ n(E_{6}) = 1

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

Probability of getting a queen of diamond = \(\frac{n\left(E_6\right)}{n(S)}=\frac{1}{52}\).

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) Total possible outcomes = 5

P(Queen card) = \(\frac{1}{5}\)

ii) If the queen card is put aside, total possible outcomes = 4.

iii) P(ace) = \(\frac{1}{4}\)

iv) P(queen) = \(\frac{0}{2}\) = 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:

Total possible outcomes = 132 + 12 = 144

No. of good pens = 132.

P(good pen) = \(\frac{132}{144}=\frac{11}{12}\)

Question 17.

i) A lot of 20 bulbs contains 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) Total possible outcomes = 20

P(Defective bulbs) = \(\frac{4}{20}=\frac{1}{5}\)

ii) Total possible outcomes = 20 – 1 = 19

No. of defective bulbs = 4

No. of good bulbs = 15

P(Not defective bulb) = \(\frac{15}{19}\)

Question 18.

A box contains 90 dises 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:

S = {1, 2, 3, 4, 5, … 90}

∴ Total possible outcomes n(S) = 90

i) Number of 2-digit numbers = 90 – 9 = 81

P(a 2-digit number) = \(\frac{81}{90}=\frac{9}{10}\)

ii) Event A = {A perfect square number}

A = (1, 4, 9, 16, 25, 36, 49, 64, 81) = n(A) = 9

Probability of the event P(A) = \(\frac{n(A)}{n(S)}=\frac{9}{90}\)

iii) A number divisible by 5, i.e., multiples of 5.

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 = 18.

P(a no. divisible by 5) = \(\frac{18}{90}=\frac{1}{5}\).

Question 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?

Solution:

Total possible outcomes = 6

No. of A’s = 2

No. of D’s = 1

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

P(D) = \(\frac{1}{6}\)

Question 20.

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:

Total number of possible outcomes = 144

No. of good pens = 144 – 20 = 124.

P(of buying) = \(\frac{124}{144}=\frac{31}{36}\)

P(of not buying) = \(\frac{20}{144}=\frac{5}{36}\)

Question 21.

(i) Two dice, one blue and one grey, are thrown at the same time. Write down all possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8, (ii) 13, (iii) less than or equal to 12? Complete the following table:

Solution:

1) Sum of 2 dice = 2 (1 + 1)

P(Sum 2) = \(\frac{1}{36}\)

2) Sum of 2 dice = 3 (1 + 2) (2 + 1)

P(Sum 3) = \(\frac{2}{36}\)

3) Sum 4 (1, 3) (2, 2) (3, 1)

P(Sum 4) = \(\frac{3}{36}\)

4) Sum 5 (1, 4) (2, 3) (3, 2) (4, 1)

P(Sum 5) = \(\frac{4}{36}\)

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

P(Sum 6) = \(\frac{5}{36}\)

6) Sum 7 (1,6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)

P(Sum 7) = \(\frac{6}{36}\)

7) Sum 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)

P(Sum 8) = \(\frac{5}{36}\)

8) Sum 9 (3, 6) (4, 5) (5, 4) (6, 3)

P(Sum 9) = \(\frac{4}{36}\)

9) Sum 10 (4, 6) (5, 5) (6, 4)

P(Sum 10) = \(\frac{3}{36}\)

10) Sum 11 (5, 6) (6,5)

P(Sum 11) = \(\frac{2}{36}\)

11) Sum 12 (6,6)

P(Sum 12) = \(\frac{1}{36}\)

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:

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 E_{1} = Sum of two dice is 3 = {(1, 2), (2, 1)}

n(E_{1}) = 2

∴ P(E_{1}) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}\)

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

n(E_{2}) = 3

∴ P(E_{2}) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{3}{36}\)

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

n(E_{3}) = 4

∴ P(E_{3}) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{4}{36}\)

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

n(E_{4}) = 5

∴ P(E_{4}) = \(\frac{5}{36}\)

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

n(E_{5}) = 6

∴ P(E_{5}) = \(\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}\)

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

n(E_{6}) = 5

∴ P(E_{6}) = \(\frac{n\left(E_6\right)}{n(S)}=\frac{5}{36}\)

g) Let E_{7} = Sum of two dice is 9 = {(3, 6), (4, 5), (5, 4), (6,3)}

n(E_{7}) = 4

∴ P(E_{7}) = \(\frac{n\left(E_7\right)}{n(S)}=\frac{4}{36}\)

h) Let E_{8} = Sum of two dice is 10 = {(4, 6), (5, 5), (6, 4)}

n(E_{8}) = 3

∴ P(E_{8}) = \(\frac{n\left(E_8\right)}{n(S)}=\frac{3}{36}\)

i) Let E_{9} = Sum of two dice is 11 = {(6,5), (5, 6)}

n(E_{9}) = 2

∴ P(E_{9}) = \(\frac{n\left(E_9\right)}{n(S)}=\frac{2}{36}\)

j) Let E_{10} = Sum of two dice is 12 = {(6, 6)}

n(E_{10}) = 1

∴ P(E_{10}) = \(\frac{n\left(E_{10}\right)}{n(S)}=\frac{1}{36}\)

No. The eleven events are not equally likely.

Question 22.

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:

Total possible outcomes (H + T)^{3}

= H^{3} + 3H^{2}T + 3HT^{2} + T^{3}

HHH HHT HTH THH HTT THT TTH TTT = 8.

Possible losses HHT HTH THH HTT THT TTH = 6

P(of losses) = \(\frac{6}{8}=\frac{3}{4}\)

Question 23.

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:

i) Total number of cases, n(S) = 6^{2} = 36

Let \(\overline{\mathrm{E}}\) = Event that 5 will come up either time

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

⇒ n(\(\overline{\mathrm{E}}\)) = 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 time = \(1-\frac{11}{36}=\frac{36-11}{36}\)

= \(\frac{25}{36}\)

ii) Probability that 5 will come up at least once = 12 – 1 = \(\frac{1}{2}\).

Question 24.

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

Solution:

i) Incorrect: We can classify the outcomes like this but they are not then ‘equally likely’. The reason is that ‘one of each’ can result in two ways – from head on first coin and tail on the second coin or from tail on the first coin and head on the second coin. This makes it twice as likely as 2 heads or 2 tails.

ii) Correct. The two outcomes considered in the question are equally likely. Both have the same probability. i.e., \(\frac{1}{2}\).