Bihar Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.

Complete the statements:

- Probability of event E + Probability of event ‘not E’= _______.
- The probability of an event that cannot happen is _______. Such an event is called _______.
- The probability of an event that is certain to happen is _______. Such an event is called _______.
- The sum of the probabilities of all the elementary events of an experiment is _______.
- The probability of an event is greater than or equal to _______ and less than or equal to _______.

Solution:

- 1
- 0, impossible event
- 1, sure or certain event
- 1
- 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 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:

(i) In the experiment “A driver attempts to start a car. The car starts or does not start”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.

(ii). In the experiment “A player attempts to shoot a basket ball. She/he shoots or misses the shot”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.

(iii) In the experiment “A trial is made to answer a true- false question. The answer is right or wrong. We know, in advance, that the result can lead in one of the two possible ways – either right or wrong. We can reasonably assume that each outcome, right or wrong, is likely to occur as the other. Thus, the outcomes right or wrong, are equally likely.

(iv) In the experiment “A baby is born. It is a boy or a girl”. We know, in advance, that the outcome can lead in one of two possible outcomes – either a boy or a girl. We are justified to assume that each outcome, a boy or a girl, is likely to occur as the other. Thus, the outcomes a boy and a girl, are equally likely.

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:

The tossing of a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game, as we know that on tossing, the coin only land in one of two possible ways – either head up or tail up. It can reasonably be assumed that each outcome, head or tail, is as likely to occur as the other, i.e., the outcomes head and tail are equally likely. So, the result of the tossing of a coin is completely 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:

Since the probability of an event E is a number P(E) such that

0 ≤ P(E) ≤ 1

∴ 1.5 cannot be the probability of an event.

∴ (B) is the correct answer.

Question 5.

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

Solution:

Since P(E) + P(not E) = 1, therefore

P(not 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) Consider the event related to the experiment of taking out of an orange flavoured candy from a bag containing only lemon flavoured candies. Since no outcome gives an orange flavoured candy, therefore it is an impossible events. So, its probability is 0.

(ii) Consider the event of taking a lemon flavoured candy out of a bag containing only lemon flavoured candies. This event is a certain event. So, its probability is 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:

Let E be the event of having the same birthday then E will be the event of 2 students not having the same birthday

i.e., P(\(\bar { E }\)) = 0.992

But P(E) + P(\(\bar { E }\)) = 1

So, P(E) = 1 – P(\(\bar { E }\))

= 1 – 0.992

= 0.008

Hence, the probability that the 2 students have 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:

There are 3 + 5 = 8 balls in a bag. Out of these 8 balls, one can be drawn in 8 ways.

∴ Total number of possible outcomes = 8

(i) Since the bag contains 3 red balls, therefore one red ball can be drawn in 3 ways. v

∴ Favourable number of outcomes to the event red ball = 3

Hence, P (getting a red ball) = \(\frac { 3 }{ 8 }\)

(ii) Since the bag contains 5 black balls along with 3 red ball, therefore one black (not red) ball can be drawn in 5 ways.

∴ Favourable number of outcomes to the event not red ball = 5

Hence, P (not getting a red ball) = \(\frac { 5 }{ 8 }\).

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 number of marbles in the box

= 5 (Red) + 8 (White) + 4 (Green)

= 17

Total number of possible outcomes = 17

(i) There are 5 red marbles in the box

∴ Favourable number of outcomes for a red marble = 5

Hence, P (getting a red marble) = \(\frac { 5 }{ 17 }\) .

(ii) There are 8 white marbles in the box.

∴ Favourable number of outcomes for a white marble = 8

Hence, P (getting a white marble) = \(\frac { 8 }{ 17 }\) .

(iii) There are 5 + 8 = 13 marbles which are not green in the box.

∴ Favourable number of outcomes for not a green marble = 13

Hence, P (not getting a green marble) = \(\frac { 13 }{ 17 }\)

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 number of coins in the piggy bank = 100 + 50 + 20 + 10 = 180

Total number of possible outcomes = 180

(i) There are one hundred 50 paise coins in the piggy bank.

∴ Favourable number of outcomes for 50 paise coin = \(\frac { 100 }{ 180 }\) = \(\frac { 5 }{ 9 }\)

(ii) There are 100 + 50 + 20, i.e., 170 coins other than Rs 5 coin.

∴ Favourable number of outcomes for not a Rs 5 coin = 170

Hence, P(falling out of a coin other than Rs 5 coin) = \(\frac { 170 }{ 180 }\) = \(\frac { 17 }{ 18 }\)

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 fishes and 8 female fishes (see figure). What is the probability that the fish taken out is a male fish?

Solution:

Total number of fishes in the tank = 5 + 8 =13

Total number of possible outcomes = 13

There are 5 male fishes in the tank.

∴ Favourable number of outcomes for a male fish = 5.

Hence, P(taking out a male 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 (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?

Solution:

Out of 8 numbers, an arrow can point any of the numbers in 8 ways.

∴ Total number of possible outcomes = 8

(i) There is only one ‘8’ on the spinning plate

∴ Favourable number of outcomes for 8 = 1

Hence, P(arrow points at 8) = \(\frac { 1 }{ 8 }\)

(ii) There are 4 odd numbers (viz. 1, 3, 5 and 7).

∴ Favourable number of outcomes for an odd number = 4.

Hence, P(arrow points at an odd number) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

(iii) There are 6 numbers greater than 2 (viz. 3, 4, 5, 6, 7 and 8).

∴ Favourable number of outcomes for a number greater than 2 = 6

Hence, P(arrow points at a number > 2) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)

(iv) There are 8 numbers less than 9 (viz. 1, 2, 3, …, 8).

∴ Favourable number of outcomes for a numberless than

Hence, P(arrow points at a number < 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: In a single throw of a die, we can get any one of the six numbers 1, 2, …, 6 marked on its six faces. Therefore, the total number of possible outcomes associated with the random experiment of throwing a die is 6.

(i) Let A denotes the event “Getting a prime number” Clearly, event A occurs, if we obtain any one of 2, 3, 5 as an outcome. Therefore,

Hence, P(a prime number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\).

(ii) Let A denotes the event “Getting a number lying between 2 and 6”. Clearly, event A occurs, if we obtain any one of 3, 4 and 5 as an outcome.

∴ Favourable number of outcomes for a number lying between 2 and 6 = 3

(iii) Let A denotes the event “Getting an odd number”. Clearly, event A occurs, if we obtain any one of 1, 3, 5 as an outcome.

∴ Favourable number of outcomes for an odd number = 3

Hence, P(an odd number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\).

Question 14.

One card is drawn from a well-shuffled deck of 52 cardjs. 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:

Out of 52 cards, one card can be drawn in 52 ways. So, total number of possible outcomes = 52.

(i) There are two suits of red cards viz. diamond and heart. Each suit contains one king.

∴ Favourable number of outcomes for a red king = 2 x 1 = 2

Hence, P(a king of red colour) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\).

(ii) In a deck of 52 cards : kings, queens and jacks are called face cards. Thus, there are 12 face cards. So, one face card can be chosen in 12 ways.

∴ Favourable number of outcomes for a face card = 12

Hence, P(a face card) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\).

(iii) There are two suits of red cards viz. diamond and heart. Each suit contains 3 face cards.

∴ Favourable number of outcomes for red face card =2 x 3 = 6

Hence, P(a red face card) = \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\).

(iv) There is only one jack of hearts

∴ Favourable number of outcomes for a jack of hearts = 1

Hence, P(the jack of hearts) = \(\frac { 1 }{ 52 }\).

(v) There are 13 cards of spade.

∴ Favourable number of outcomes for a spade = 13

Hence, P(a spade) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\).

(vi) There is only one queen of diamonds.

∴ Favourable number of outcomes for a queen of diamonds = 1

Hence, P(the queen of diamonds) = \(\frac { 1 }{ 52 }\).

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? (6) a queen?

Solution:

Out of five cards – the ten, jack, queen, king and ace of diamonds, one card can be drawn in 5 ways. So, total number of possible outcomes = 5

(i) There is only one queen.

∴ Favourable number of outcomes for a queen = 1

Hence, P(the queen) = \(\frac { 1 }{ 5 }\).

(ii) After keeping the queen card aside, we are left with 4 cards. So, total number of possible ou tcomes now = 4.

(a) There is only one ace

∴ Favourable number of outcomes for an ace = 1 1

Hence, Plan ace) = \(\frac { 1 }{ 4 }\).

(b) There is no card as queen.

∴ Favourable number of outcomes for a queen = 0

Hence, P(the queen) = \(\frac { 0 }{ 4 }\) = 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:

Out of 132 + 12, i.e., 144 pens, one pen can be chosen in 144 ways.

∴ Total number of possible outcomes = 144

There are 132 non-defective, i.e., good pens, out of which one pen can be chosen in 132 ways.

∴ Favourable number of outcomes for a good pen = 132

Hence, P(getting a 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) Out of 20 bulbs one bulb can be chosen in 20 ways.

∴ Total number of possible outcomes = 20 There are 4 defective bulbs out of which one bulb can be chosen in 4 ways.

So, favourable number of outcomes for a defective bulb = 4.

Hence, P(getting a defective bulb) = \(\frac { 4 }{ 20 }\) = \(\frac { 1 }{ 5 }\)

(ii) On drawing a non-defective bulb out of 20 bulbs, we are left with 19 bulbs including 4 defective bulbs.

∴ Total number of possible outcomes = 19

There are 19 – 4 = 15 non-defective bulbs, out of which one bulb can be drawn in 15 ways.

∴ Favourable number of outcomes for a non-defective bulb = 15

Hence, P(getting a non-defective bulb) = \(\frac { 15 }{ 19 }\).

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 discs bearing numbers 1 to 90 in the box of which one disc can be drawn in 90 ways.

∴ Total number of possible outcomes = 90

(i) There are 90 – 9, i.e., 81 discs bearing a two-digit number in the box of which one disc can be drawn in 81 ways.

∴ Favourable number of outcomes for a 2-digit number = 81

Hence, P(getting a disc bearing a two-digit number) = \(\frac { 81 }{ 90 }\) = \(\frac { 9 }{ 10 }\).

(ii) Those numbers from 1 to 90 which are perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, i.e., squares of 1, 2, 3, 4, 5, 6, 7, 8 and 9 respectively.

Therefore, there are 9 discs marked with numbers which are perfect squares.

∴ Favourable number of outcomes for a perfect square = 9

Hence, PCgetting a disc marked with a number which is a perfect square)

= \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)

(iii) Those numbers from 1 to 90 which are divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90. They are 18 in number.

Therefore, there are 18 discs marked with the numbers which are divisible by 5.

Favourable number of outcomes for divisible by 5 = 18

Hence, P(getting a disc marked with a number which is 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:

In a single throw of a die, we can get any one of the six letters A, B, C, D, E and A marked on its faces. Therefore, the total numbers of possible outcomes associated with the random experiment of throwing a die is 6.

(i) Let X be the event “Getting a letter A”. Clearly, event X occurs on the two faces. –

∴ Favourable number of possible outcomes for A = 2

∴ p(X) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

(ii) Let Y be the event “Getting a letter D”. Cleanly event Y occurs on one face.

∴ Favourable number of outcomes for D = 1

∴ p(Y) = \(\frac { 1 }{ 6 }\)

Question 20.

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

Solution:

Total area of the figure, i.e., rectangle

= 3 m x 2 m = 6 m²

Area of the circle = πr² = π(\(\frac { 1 }{ 2 }\)m)² = \(\frac { π }{ 4 }\) m²

Probability (die to land inside the circle)

= \(\frac { π/4 }{ 6 }\)

= \(\frac { π }{ 24 }\)

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:

There is a lot of 144 ball pens. One pen can be drawn in 144 ways so total number of possible outcomes = 144. Out of these 144 ball pens 20 are defective ball pens.

Number of non-defective ball pens

= 144 – 20 = 124

(i) P(she will buy) = P(a non-defective pen)

= \(\frac { 124 }{ 144 }\) = \(\frac { 31 }{ 36 }\)

(ii) P(she will not buy) = P(a defective pen)

= \(\frac { 20 }{ 144 }\) = \(\frac { 5 }{ 36 }\)

Question 22.

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

(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:

Possible outcomes associated to the random experiment of throwing two dice are –

(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).

∴ Total number of possible outcomes = 6 x 6 = 36

Let A be the event of getting the sum as 3.

The outcomes favourable to event A are (1, 2) and (2, 1)

Clearly, favourable number of outcomes for sum 3 = 2

Hence, required probability = \(\frac { 2 }{ 36 }\)

Let A be the event of getting the sum as 4.

The outcomes favourable to event A are (1, 3), (3, 1) and (2, 2)

Clearly, favourable number of outcomes for sum 4 = 3

Hence, required probability = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)

Let A be the event of getting the sum as 5.

The outcomes favourable to event A are (1, 4), (4, 1), (2, 3) and (3, 2)

Clearly, favourable number of outcomes for sum 5 = 4

Hence, required probability = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)

Let A be the event of getting the sum as 6.

The outcomes favourable to event A are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3).

Clearly, favourable number of outcomes for sum 6 = 5

Hence, required probability = \(\frac { 5 }{ 36 }\)

Let A be the event of getting the sum as 7.

The outcomes favourable to event A are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4, 3).

Clearly, favourable number of outcomes for sum 7 = 6

Hence, required probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\).

Let A be the event of getting the sum as 9.

The outcomes favourable to event A are (3, 6), (6, 3), (4, 5) and (5, 4).

Clearly, favourable number of outcomes for sum 9=4

Hence, required probability = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)

Let A be the event of getting the sum as 10.

The outcomes favourable to event A are (4, 6), (6, 4) and (5, 5).

Clearly, favourable number of outcomes for sum 10 = 3

Hence, required probability = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)

Let A be the event of getting the sum as 11.

The outcomes favourable to event A are (5, 6) and (6, 5). Clearly, favourable number of outcomes for sum 11 = 2

Hence, required probability = \(\frac { 2 }{ 36 }\) = \(\frac { 1 }{ 18 }\)

Thus, the complete table is as on the next page :

(ii) I do not agree with the argument given here. Justification has already been given in part (i).

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:

Consider the experiment in which a coin is tossed thrice. The outcomes associated with this experiment are given by HHH, HHT, HTH, THH, TTH, HTT, THT, TTT.

∴ Total number of possible outcomes = 8.

Hanif will lose the game, if he gets HHT, HTH, THH, TTH, HTT, THT.

∴ Favourable number of outcomes for Hanif losing = 6

Hence, required probability = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)

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?

Solution:

(i) See the table of 36 possible outcomes of question 22 above, 5 is not appearing in following : (1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6).

There are 25 such outcomes favourable to not 5.

So, P(not 5) = \(\frac { 25 }{ 36 }\)

(ii) Again refer to the same table of question 22. Favourable outcomes to 5 coming at least once are : (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).

They are 11 such outcomes.

So, P(at least once 5) = \(\frac { 11 }{ 36 }\)

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 \(\frac { 3 }{ 36 }\).

(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’. Reason is that ‘one of each’ can result in two ways – from a head on first coin and tail on the second coin or from a tail on the first coin and head on the second coin. This makes it twice as likely as two heads (or two tails).

(ii) Correct. The two outcomes considered in the question are equally likely.