Class 10 - Chapter 14: Probability

Exercise 14.1

Q1. Complete the following statements:

(i)

Probability of an Event E + Probability of the event 'not E' = 1

(ii)

The probability of an event that cannot happen is 0.

Such an event is called an impossible event.

(iii)

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

Such an event is called a sure (certain) event.

(iv)

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 0 and less than or equal to 1.

Q2. 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.

The two outcomes are not equally likely because the car is usually more likely to start than not start.

Answer: Not equally likely.

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

The outcomes depend on the player's skill.

Answer: Not equally likely.

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

If the answer is chosen randomly, both outcomes have equal chances.

Answer: Equally likely.

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

Both outcomes are generally considered equally likely.

Answer: Equally likely.

Q3. 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?

A coin has two outcomes: Head and Tail.

Both outcomes are equally likely.

Therefore, each team has an equal chance of winning the toss.

Hence, tossing a coin is considered a fair method of making the decision.

Q4. Which of the following cannot be the probability of an event?

(A) 2/3

(B) −1.5

(C) 15% = 15/100 = 0.15

(D) 0.7

Since probability always lies between 0 and 1,

−1.5 cannot be a probability.

Answer: (B) −1.5

Q5. If P(E) = 0.05, what is the probability of 'not E'?

P(not E) = 1 − P(E)

= 1 − 0.05

= 0.95

Answer: 0.95

Q6. 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?

Since there are no orange flavoured candies in the bag,

P(orange candy) = 0

Answer: 0

(ii) a lemon flavoured candy?

All candies are lemon flavoured.

P(lemon candy) = 1

Answer: 1

Q7. 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?

P(same birthday)

= 1 − P(not same birthday)

= 1 − 0.992

= 0.008

Answer: 0.008

Q8. 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:

Total number of balls = 3 + 5 = 8

(i) Red

P(red)

= Number of red balls / Total balls

= 3/8

Answer: 3/8

(ii) Not red

P(not red)

= Number of black balls / Total balls

= 5/8

Answer: 5/8

Q9. 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:

Total number of marbles

= 5 + 8 + 4

= 17

(i) Red

P(red)

= Number of red marbles / Total marbles

= 5/17

Answer: 5/17

(ii) White

P(white)

= Number of white marbles / Total marbles

= 8/17

Answer: 8/17

(iii) Not Green

Number of marbles which are not green

= 5 + 8

= 13

P(not green)

= 13/17

Answer: 13/17

Q10. 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:

Total number of coins

= 100 + 50 + 20 + 10

= 180

(i) will be a 50p coin

P(50p coin)

= 100/180

= 5/9

Answer: 5/9

(ii) will not be a ₹5 coin

Number of coins which are not ₹5 coins

= 180 − 10

= 170

P(not ₹5 coin)

= 170/180

= 17/18

Answer: 17/18

Q11. 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?

Number of male fish = 5

Number of female fish = 8

Total fish = 5 + 8 = 13

P(male fish)

= Number of male fish / Total fish

= 5/13

Answer: 5/13

Q12. 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:

Total possible outcomes = 8

S = {1, 2, 3, 4, 5, 6, 7, 8}

(i) 8

Favourable outcomes = {8}

P(8)

= 1/8

Answer: 1/8

(ii) An odd number

Odd numbers = {1, 3, 5, 7}

Number of favourable outcomes = 4

P(odd number)

= 4/8

= 1/2

Answer: 1/2

(iii) A number greater than 2

Numbers greater than 2

= {3, 4, 5, 6, 7, 8}

Number of favourable outcomes = 6

P(number greater than 2)

= 6/8

= 3/4

Answer: 3/4

(iv) A number less than 9

Numbers less than 9

= {1, 2, 3, 4, 5, 6, 7, 8}

Number of favourable outcomes = 8

P(number less than 9)

= 8/8

= 1

Answer: 1

Q13. A die is thrown once. Find the probability of getting:

Sample Space:

S = {1, 2, 3, 4, 5, 6}

Total outcomes = 6

(i) A Prime Number

Prime numbers on a die = {2, 3, 5}

Favourable outcomes = 3

P(prime number)

= 3/6

= 1/2

Answer: 1/2

(ii) A Number Lying Between 2 and 6

Numbers between 2 and 6 = {3, 4, 5}

Favourable outcomes = 3

P(number between 2 and 6)

= 3/6

= 1/2

Answer: 1/2

(iii) An Odd Number

Odd numbers = {1, 3, 5}

Favourable outcomes = 3

P(odd number)

= 3/6

= 1/2

Answer: 1/2

Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

Total number of cards = 52

(i) A King of Red Colour

Red kings = King of Hearts, King of Diamonds

Favourable outcomes = 2

P(red king)

= 2/52

= 1/26

Answer: 1/26

(ii) A Face Card

Face cards = J, Q, K

Number of face cards = 12

P(face card)

= 12/52

= 3/13

Answer: 3/13

(iii) A Red Face Card

Red face cards

= 2 red suits × 3 face cards

= 6

P(red face card)

= 6/52

= 3/26

Answer: 3/26

(iv) The Jack of Hearts

Only one such card exists.

P(Jack of Hearts)

= 1/52

Answer: 1/52

(v) A Spade

Number of spade cards = 13

P(spade)

= 13/52

= 1/4

Answer: 1/4

(vi) The Queen of Diamonds

Only one such card exists.

P(Queen of Diamonds)

= 1/52

Answer: 1/52

Q15. Five cards — the Ten, Jack, Queen, King and Ace of Diamonds — are well-shuffled with their face downwards. One card is picked up at random.

Total cards = 5

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

Favourable outcomes = 1

P(Queen)

= 1/5

Answer: 1/5

(ii) If the Queen is drawn and put aside, what is the probability that the second card picked up is:

(a) an Ace?

Remaining cards = 4

Number of aces = 1

P(Ace)

= 1/4

Answer: 1/4

(b) a Queen?

The Queen has already been removed.

P(Queen)

= 0/4

= 0

Answer: 0

Q16. 12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Number of good pens = 132

Number of defective pens = 12

Total pens = 144

P(good pen)

= 132/144

= 11/12

Answer: 11/12

Q17.

(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?

Total bulbs = 20

Defective bulbs = 4

P(defective bulb)

= 4/20

= 1/5

Answer: 1/5

(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?

Initially:

Good bulbs = 16

Defective bulbs = 4

One good bulb has already been removed.

Remaining good bulbs = 15

Remaining total bulbs = 19

P(not defective)

= 15/19

Answer: 15/19

Q18. A box contains 90 discs numbered from 1 to 90. One disc is drawn at random. Find the probability that it bears:

Total outcomes = 90

(i) A Two-Digit Number

Two-digit numbers from 10 to 90

Count = 81

P(two-digit number)

= 81/90

= 9/10

Answer: 9/10

(ii) A Perfect Square Number

Perfect squares ≤ 90:

1, 4, 9, 16, 25, 36, 49, 64, 81

Count = 9

P(perfect square)

= 9/90

= 1/10

Answer: 1/10

(iii) A Number Divisible by 5

Numbers divisible by 5:

5, 10, 15, ..., 90

Count = 90/5 = 18

P(number divisible by 5)

= 18/90

= 1/5

Answer: 1/5

Q19. A child has a die whose six faces show the letters A, B, C, D, E, A. The die is thrown once. What is the probability of getting:

Faces of the die:

A, B, C, D, E, A

Total outcomes = 6

(i) A

'A' appears on 2 faces.

P(A)

= 2/6

= 1/3

Answer: 1/3

(ii) D

'D' appears on 1 face.

P(D)

= 1/6

Answer: 1/6

Q20. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1 m?

Length of rectangle = 3 m

Breadth of rectangle = 2 m

Area of rectangle

= 3 × 2

= 6 m²

Diameter of circle = 1 m

Radius = 0.5 m

Area of circle

= πr²

= (22/7) × (0.5)²

= 11/14 m²

Probability

= Area of circle / Area of rectangle

= (11/14)/6

= 11/84

Answer: 11/84

Q21. 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 it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:

Total pens = 144

Defective pens = 20

Good pens = 144 − 20 = 124

(i) She will buy it

P(good pen)

= 124/144

= 31/36

Answer: 31/36

(ii) She will not buy it

P(defective pen)

= 20/144

= 5/36

Answer: 5/36

Q22. Refer to Example 13(i).

(i) Complete the following table:

When two dice are thrown,

Total outcomes = 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 probability 1/11. Do you agree?

No.

The sums are not equally likely.

For example:

Sum 7 can occur in 6 ways:

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

But sum 2 can occur in only 1 way:

(1,1)

Therefore, all sums do not have equal probabilities.

Answer: No, the argument is incorrect.

Q23. 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.

Total outcomes

= 2³

= 8

Sample space:

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Hanif wins when:

HHH or TTT

Number of winning outcomes = 2

P(win)

= 2/8

= 1/4

P(lose)

= 1 − 1/4

= 3/4

Probability that Hanif loses = 3/4

Q24. A die is thrown twice. What is the probability that:

Total outcomes = 6 × 6 = 36

(i) 5 will not come up either time

Probability of not getting 5 in one throw

= 5/6

Probability of not getting 5 in both throws

= (5/6) × (5/6)

= 25/36

Answer: 25/36

(ii) 5 will come up at least once

P(at least one 5)

= 1 − P(no 5 in both throws)

= 1 − 25/36

= 11/36

Answer: 11/36