# PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1

PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1

## PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.

Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = _________.

Solution:

Probability of an event E +

Probability of the event ‘not E’ = 1

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

Solution:

The probability of an event that cannot happen is 0. Such an event is called impossible event.

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

Solution:

The probability of an event that is certain to happen is 1. Such event is called sure event.

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

Solution:

The sum of the probabilities of all the elementary events of an experiment is 1.

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

Solution:

The probability of an event is greater than or equal to 0 and less than or equal to 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.

Solution:

When a dnver attempts to start a car the car starts normally. Only when there is some defects the car does not start. So the outcome is not equally likely.

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

Solution:

When a player attempts to shoot a basketball the outcome in this situation is not equally likely because the outcome depends on many factors such as the training of the player, quality of the gun used etc.

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

Solution:

Since for a question there are two possibilities either right or wrong the outcome in this trial of true-false question is either true or false i.e. one out of the two and both have equal chances to happen. Hence, the two outcomes are equally likely.

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

Solution:

A new baby (i.e. who took birth at a moment) can be either a boy or a girl and both the outcome have equally likely chances.

Question 3.

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

Solution:

When a coin is tossed there are only two possibilities i.e. Head or tail both are equally likely to happen. Result of the toss of a fair coin is completely unpredictable.

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) Since bag contains only lemon flavoured candies

∴ There is no orange candies

∴ It is impossible event.

∴ Probability of getting orange flavoured = 0.

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:

Number of red marbles = 5

Number of white marbles = 8

Number of green marbles = 4

Total number of marbles = 5 + 8 + 4 = 17

Since, one marble is taken out

(i) There are 5 Red marbles

Question 10.

A piggy bank contains hundred 50p 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;

Number of 50 coins = 100

Number of ₹ 1 coins = 50

Number of ₹ 2 coins =20

Number of ₹ 5 coins = 10

∴ Total number of coins = 100 + 50 + 20 + 10 = 180

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?

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?

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:

When dice is thrown number of possible outcomes

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

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:

There are 52 cards in a pack

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:

Five cards are ten, jack, queen, king and ace

(i) Probability of getting queen = 1/5

∴ P (A queen) = 1/5.

(ii) If the queen is drawn and put aside then there are 4 cards left – Ten, a Jack, a king and an ace.

(a) Probability of getting an ace = 1/4

P (An Ace) = 1/4.

There’s no queen left

(b) Probability of getting a queen = 0/4 = 0

P (a queen) = 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:

Number of defective pens = 12

Number of good pens = 132

∴ Total number of pens = 12 + 132 = 144

Probability of getting good pen = 132/144=11/12

P (a good pen) = 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 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) Number of defective bulbs 4

Number of good bulbs (Not defective) = 16

Total number of bulbs = 4 + 16 = 20

Probability of getüng defective bulb = 4/20=1/5.

(ii) When a defective bulb drawn is not being replaced, we are left with 19 bulbs

Now probability of getting not defective bulb = 15/19

∴ P (Not defective bulb) = 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:

From 1 to 90 there are 90 numbers in all and 81 two – digit numbers from 10 to 90

(i) Probability of getting two digit number = 81/90

∴ P (two digit number) = 81/90 = 9/10.

(ii) Perfect square numbers are (1, 4, 9, 16, 25, 36, 49, 64, 81 } there are 9 perfect square numbers between 1 to 90

Probability of getting perfect square = 9/90=1/10

∴ P (Perfect square) = 1/10

(iii) Numbers divisible by 5 are (5, 10, 15, 20, 25, 30, 35, 40, 45. 50, 55. 60, 65, 70, 75, 80, 85, 90}

There are 18 numbers divisible by 5

∴ Probability of number getting divisible by 5 = 18/90=1/5

∴ Required probability = 1/5.

Question 19.

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

Question 20.

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

Question 21.

A lot consists of 144 ball pens of which 20 are défective and the others are good. Nun will buy a pen if it is good, but will not buy it 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 Pens in lot = 144

Number of defective Pens = 20

∴ Number of good Pens = 144 – 20 = 124

Question 22.

Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes

(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 1/11 Do you agree with this argument ? Justify your answer.

Solution:

When two dices are thrown total number of possible outcomes

{(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)

S (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

Let A is event of getting sum as 3

∴ A = {(1,2) (2, 1)}

n(A) = 2

∴ Probability of getting sum as 3 = 2/36 = 1/18

P(A) = 1/18

(ii) No, here all 11 possible outcomes are not equally likely

∴ Three probabilites are different.

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:

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:

When a die is thrown twice all possible outcomes are

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) 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

Ler A is event that 5 will come up either time

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

n(A) = 11

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