MCQs for Probability

Below are the MCQs for Chapter 15-Probability.

Question. The probability of event equal to zero is called;

(a) Unsure event

(b) Sure Event

(c) Impossible event

(d) Independent event

Answer: (c) Impossible event

Explanation: The probability of an event that cannot happen or which is impossible, is equal to zero.

Question. The probability that cannot exist among the following:

(a) ⅔

(b) -1.5

(c) 15%

(d) 0.7

Answer: (b) -1.5

Explanation: The probability lies between 0 and 1. Hence, it cannot be negative.

Question. If P(E) = 0.07, then what is the probability of ‘not E’?

(a) 0.93

(b) 0.95

(c) 0.89

(d) 0.90

Answer: (a) 0.93

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

Since, P(E) = 0.05

So, P(not E) = 1 – P(E)

Or, P(not E) = 1 – 0.07

∴ P(not E) = 0.93

Question. A bag has 3 red balls and 5 green balls. If we take a ball from the bag, then what is the probability of getting red balls only?

(a) 3

(b) 8

(c) ⅜

(d) 8/3

Answer: (c) ⅜

Explanation: Number of red balls = 3

Number of green balls = 5

Total balls in bag = 3+5 = 8

Probability of getting red balls = number of red balls/total number of balls

= ⅜

Question. A bag has 5 white marbles, 8 red marbles and 4 purple marbles. If we take a marble randomly, then what is the probability of not getting purple marble?

(a) 0.5

(b) 0.66

(c) 0.08

(d) 0.77

Answer: (d) 0.77

Explanation: Total number of purple marbles = 4

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

Probability of getting purple marbles = 4/17

Hence, the probability of not getting purple marbles = 1-4/17 = 0.77

Question. A dice is thrown in the air. The probability of getting odd numbers is

(a) ½

(b) 3/2

(c) 3

(d) 4

Answer: (a) ½

Explanation: A dice has six faces having values 1, 2, 3, 4, 5 and 6.

There are three odd numbers and three even numbers.

Therefore, the probability of getting only odd numbers is = 3/6 = ½

Question. If we throw two coins in the air, then the probability of getting both tails will be:

(a) ½

(b) ¼

(c) 2

(d) 4

Answer: (b) ¼

Explanation: When two coins are tossed, the total outcomes will be = 2 x 2 = 4

Hence, the probability of getting both tails = ¼

Question. If two dice are thrown in the air, the probability of getting sum as 3 will be

(a) 2/18

(b) 3/18

(c) 1/18

(d) 1/36

Answer: (c) 1/18

Explanation: When two dice are thrown in the air:

Total number of outcome = 6 x 6 = 36

Sum 3 is possible if we get (1,2) or (2,1) in the dices.

Hence, the probability will be = 2/36 = 1/18

Question. A card is drawn from the set of 52 cards. Find the probability of getting a queen card.

(a) 1/26

(b) 1/13

(c) 4/53

(d) 4/13

Answer: (b) 1/13

Explanation: Total number of cards = 52

Number of queen cards= 4

The probability of getting queen card = 4/52 = 1/13

10. A fish tank has 5 male fish and 8 female fish. The probability of fish taken out is a male fish:

(a) 5/8

(b) 5/13

(c) 13/5

(d) 5

Answer: (b) 5/13

Explanation: Total fish = 5 + 8 = 13

Probability of taking out a male fish = 5/13

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

(a) 0.5

(b) 1

(c) 2

(d) 1.5

Answer: (b) 1

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

Question. A card is selected at random from a well shuffled deck of 52 playing cards. The probability of its being a face card is

(a) 3/13

(b) 4/13

(c) 6/13

(d) 9/13

Answer: (a) 3/13

Explanation:

Total number of outcomes = 52

Number of face cards = 12

The probability of its being a face card = 12/52 = 3/13

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

(a) 1

(b) 3/4

(c) 1/2

(d) 0

Answer: (d) 0

If an event cannot occur, then its probability is 0.

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

(a) 0.0001

(b) 0.001

(c) 0.01

(d) 0.1

Answer: (a) 0.0001

The probability of an event which is very unlikely to happen is closest to zero.

Thus, 0.0001 is the probability of an event which is very unlikely to happen.

Question. If P(A) denotes 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

Answer: (c) 0 ≤ P(A) ≤ 1

If P(A) denotes the probability of an event A, then 0 ≤ P(A) ≤ 1.

Question. The probability that a non leap year selected at random will contain 53 Sundays is

(a) 1/7

(b) 2/7

(c) 3/7

(d) 5/7

Answer: (a) 1/7

Explanation:

Non-leap year = 365 days

365 days = 52 weeks + 1 day

For 52 weeks, number of Sundays = 52

1 remaining day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.

Total possible outcomes = 7

The number of favourable outcomes = 1

Thus, the probability of getting 53 Sundays = 1/7

Question. If the probability of an event is p, the probability of its complementary event will be

(a) p – 1

(b) p

(c) 1 – p

(d) 1 – 1/p

Answer: (c) 1 – p

Explanation:

The sum of probability of an event and it complementary event = 1

So, if the probability of an event is p, the probability of its complementary event will be 1 – p.

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

(a) 4

(b) 13

(c) 48

(d) 51

Answer: (d) 51

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

Given that the event E is that card is not an ace of hearts.

Hence, the number of outcomes favorable to E = 52 – 1 = 51

Question. 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

Answer: (b) 14

Explanation:

Total number of eggs = 400

Probability of getting a bad egg = Number of bad eggs/Total number of eggs

0.035 = Number of bad eggs/400

Number of bad eggs = 0.035 × 400 = 14

Question. Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. The probability of Reshma winning the match is

(a) 0.62

(b) 0.38

(c) 0.58

(d) 0.42

Answer: (b) 0.38

Explanation:

Probability of Sangeeta’s winning = P(S) = 0.62

Probability of Reshma’s winning = P(R) = 1 – P(S) {since events R and S are complementary}

= 1 – 0.62

= 0.38.

Question.  The probability of getting exactly one head in tossing a pair of coins is

(a) 0

(b) 1

(c) 1/3

(d)1/2

Question.  A dice is thrown. Find the probability of getting an even number.

(a) 2/3

(b) 1

(c) 5/6

(d) 1/2

Answer: (d) 1/2

Explanation: Total number of cases = 6 (1,2,3,4,5,6)

There are three even numbers 2,4,6

Therefore probability of getting an even number is:

P (even) = 3/6

⇒ P (even) = 1/2

Question.  Two coins are thrown at the same time. Find the probability of getting both heads.

(a) 3/4

(b) 1/4

(c) 1/2

(d) 0

Answer: (b) 1/4

Explanation: Since two coins are tossed, therefore total number of cases = 2² = 4

Therefore, probability of getting heads in both coins is:

∴ P (head) = 1/4

Question. Two dice are thrown simultaneously. The probability of getting a sum of 9 is:

(a) 1/10

(b) 3/10

(c) 1/9

(d) 4/9

Answer: (c) 1/9

Explanation: Total cases = 36

Total cases in which sum of 9 can be obtained are:

(5, 4), (4, 5), (6, 3), (3, 6)

∴ P (9) = 4/36 = 1/9

Question. 100 cards are numbered from 1 to 100. Find the probability of getting a prime number.

(a) 3/4

(b) 27/50

(c) 1/4

(d) 29/100

Answer: (c) 1/4

Explanation: Total prime numbers from 1 to 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

That means 25 out of 100

So probability is:

P (prime) = 25/100

⇒ P (prime) = 1/4

Question. 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, then the number of blue balls in a bag is:

(a) 5

(b) 10

(c) 15

(d) 20

Answer: (b) 10

Explanation: Let the number of blue balls be x

Then total number of balls will be 5 + x.

According to question,

x/(5 + x) = 2 X (5/5+x)

⇒ x = 10

Question. A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

(a) 143/150

(b) 147/150

(c) 1/25

(d) 1/50

Answer: (b) 147/150

Explanation:

P (non-defective bulb) = 1 – P (Defective bulb)

= 1 – (12/600)

= (600 – 12)/600

= 588/600

= 147/150

Question.  Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box randomly, then the probability that the number on card is a perfect square.

(a) 9/100

(b) 1/10

(c) 3/10

(d) 19/100

Answer: (b) 1/10

Explanation: The perfect square numbers between 2 to 101 are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Total numbers from 2 to 101 =100

So probability of getting a card with perfect square number is:

P (perfect square) = 10/100

⇒ P (perfect square) = 1/10

Question.  What is the probability of getting 53 Mondays in a leap year?

(a) 1/7

(b) 53/366

(c) 2/7

(d) 7/366

Answer: (c) 2/7

Explanation: With 366 days, the number of weeks in a year is

366/7 = 52 (2/7)

i.e., 52 complete weeks which contains 52 Mondays,

Now 2 days of the year are remaining.

These two days can be

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

i.e., there are 7 pairs, in which Monday occurs in 2 pairs,

So probability is:

P (53 Monday) = 2/7

Question.  A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a king of red suit.

(a) 1/26

(b) 3/26

(c) 7/52

(d) 1/13

Answer: (a) 1/26

Explanation: There are total 4 kings in 52 cards, 2 of red colour and 2 of black colour

Therefore, Probability of getting a king of red suit is:

P (King of red suit) = 2/52

⇒ P (King of red suit) = 1/26

Question. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1,2,3……12 ,then the probability that it will point to an odd number is:

(a) 1/6

(b) 1/12

(c) 7/12

(d) 5/12

Answer: (a) 1/6

Explanation: The odd numbers in 1,2,3……..12 are:

1,3,5,7,9,11

Therefore probability that an odd number will come is:

P (odd number) = 6/12

⇒ P (odd number) = 1/2

Question.  A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Aryan wins if all the tosses give the same result i.e. three heads or three tails and loses otherwise. Then the probability that Aryan will lose the game.

(a) 3/4

(b) 1/2

(c) 1

(d) 1/4

Answer: (a) 3/4

Explanation: Total outcomes are:

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

Favourable outcomes for losing game are

HHT, HTH, THH, HTT, THT, TTH

Therefore probability of losing the game is:

P (Losing the game) = 6/8

⇒ P (Losing the game) = 3/4

Question. Riya and Kajal are friends. Probability that both will have the same birthday isthe same birthday is:

(a) 364/365

(b) 31/365

(c) 1/365

(d) 1/133225

Answer: (c) 1/365

Explanation:

Riya may have any one of 365 days of the year as her birthday. Similarly Kajal may have any one of 365days as her birthday.

Total number of ways in which Riya and Kajal may have their birthday are:

365 × 365

Then Riya and Kajal may have same birthday on any one of 365 days.

Therefore number of ways in which Riya and Kajal may have same birthday are:

= 365/365 X 365

= 1/365

Question.  A number x is chosen at random from the numbers -2, -1, 0 , 1, 2. Then the probability that x² < 2 is?

(a) 1/5

(b) 2/5

(c) 3/5

(d) 4/5

Answer: (c) 3/5

Explanation: We have 5 numbers −2,−1,0,1,2

Whose squares are 4,1,0,1,4

So square of 3 numbers is less than 2

Therefore Probability is:

P (x² < 2) = 3/5

Question.  A jar contains 24 marbles. Some are red and others are white. If a marble is drawn at random from the jar, the probability that it is red is 2/3, then the number of white marbles in the jar is:

(a) 10

(b) 6

(c) 8

(d) 7

Answer: (c) 8

Explanation: Let the number of white marbles be x.

Since only two colour marbles are present, and total probability we know of all the events is equal to 1.

P (white) = 1 – P (red)

x/24 = 1 – (2/3)

⇒ x/24 = 1/3

⇒ x = 8

So there are 8 white marbles.

Question.  A number is selected at random from first 50 natural numbers. Then the probability that it is a multiple of 3 and 4 is:

(a) 7/50

(b) 4/25

(c) 1/25

(d) 2/25

Answer: (d) 2/25

Explanation: The numbers that are multiple of 3(from first 50 natural numbers) are:

3, 6, 9, 12, 15, 18………………..48

The numbers that are multiple of 4 (from first 50 natural numbers) are:

4, 8, 12, 16…………………….48

The numbers that are multiples of 3 and 4 both are the multiples of 3×4=12 as both 3 and 4 are co-prime.

So common multiples are:

12, 24, 36, 48

Therefore probability is:

P (multiple of 3 and 4) = 4/50

⇒ P (multiple of 3 and 4) = 2/25