Probability — Multiple Choice Questions

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

1. 0.7

2. $\dfrac{2}{3}$

3. -1.5

4. 15%

Answer

Since, the probability cannot be negative.

Hence, Option 3 is the correct option.

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

1. p - 1

2. p

3. 1 - p

4. 1 - $\dfrac{1}{p}$

Answer

Probability of an event is p.

∴ Probability of complementary event is 1 - p.

Hence, Option 3 is the correct option.

Out of one digit prime numbers, one number is selected at random. The probability of selecting an even number is

1. $\dfrac{1}{2}$

2. $\dfrac{1}{4}$

3. $\dfrac{4}{9}$

4. $\dfrac{2}{5}$

Answer

Sample space = {2, 3, 5, 7}.

Let A be the event of selecting an even number,

∴ A = {2}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\dfrac{1}{4}.$

Hence, Option 2 is the correct option.

When a die is thrown, the probability of getting an odd number less than 3 is

1. $\dfrac{1}{6}$

2. $\dfrac{1}{3}$

3. $\dfrac{1}{2}$

4. 0

Answer

When a die is thrown the sample space is,

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

Let A be the event of getting an odd number less than 3,

∴ A = {1}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\dfrac{1}{6}.$

Hence, Option 1 is the correct option.

The probability of getting a number divisible by 3 in throwing a die is

1. $\dfrac{1}{6}$

2. $\dfrac{1}{3}$

3. $\dfrac{1}{2}$

4. $\dfrac{2}{3}$

Answer

There are 2 numbers between 1 to 6 that are divisible by 3, i.e., 3 and 6.

∴ No. of favourable outcomes = 2

Total no. in a dice = 6.

∴ No. of possible outcomes = 6

P(a number divisible by 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$.

Hence, option 2 is the correct option.

A fair die is thrown once. The probability of getting an even prime number is

1. $\dfrac{1}{6}$

2. $\dfrac{2}{3}$

3. $\dfrac{1}{3}$

4. $\dfrac{1}{2}$

Answer

When a die is thrown the sample space is,

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

Let A be the event of getting an even prime number,

∴ A = {2}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\dfrac{1}{6}.$

Hence, Option 1 is the correct option.

A fair die is thrown once. The probability of getting a composite number is

1. $\dfrac{1}{3}$

2. $\dfrac{1}{6}$

3. $\dfrac{2}{3}$

4. 0

Answer

When a die is thrown the sample space is,

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

Let A be the event of getting a composite number,

∴ A = {4, 6}.

Hence, the no. of favourable outcomes to A = 2.

∴ P(A) = $\dfrac{2}{6} = \dfrac{1}{3}.$

Hence, Option 1 is the correct option.

If a fair die is rolled once, then the probability of getting an even number or a number greater than 4 is

1. $\dfrac{1}{2}$

2. $\dfrac{1}{3}$

3. $\dfrac{5}{6}$

4. $\dfrac{2}{3}$

Answer

When a die is thrown the sample space is,

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

Let A be the event of getting an even number or a number greater than 4,

∴ A = {2, 4, 5, 6}.

Hence, the no. of favourable outcomes to A = 4.

∴ P(A) = $\dfrac{4}{6} = \dfrac{2}{3}.$

Hence, Option 4 is the correct option.

If a letter is chosen at random from the letters of English alphabet, then the probability that it is a letter of the word 'DELHI' is

1. $\dfrac{1}{5}$

2. $\dfrac{1}{26}$

3. $\dfrac{5}{26}$

4. $\dfrac{21}{26}$

Answer

Total no. of letters in English alphabet = 26.

Let A be the event of getting a letter of the word 'DELHI',

∴ A = {D, E, L, H, I}.

Hence, the no. of favourable outcomes to A = 5.

∴ P(A) = $\dfrac{5}{26}.$

Hence, Option 3 is the correct option.

A card is selected at random from a pack of 52 cards. The probability of its being a red face card is

1. $\dfrac{3}{26}$

2. $\dfrac{3}{13}$

3. $\dfrac{2}{13}$

4. $\dfrac{1}{2}$

Answer

There are 6 red face cards, 3 of hearts and 3 of diamonds.

∴ The number of favourable outcomes to event 'red face card' = 6.

∴ P(red face card) = $\dfrac{6}{52} = \dfrac{3}{26}$.

Hence, Option 1 is the correct option.

If a card is drawn from a well-shuffled pack of 52 playing cards, then the probability of this card being a king or jack is

1. $\dfrac{1}{26}$

2. $\dfrac{1}{13}$

3. $\dfrac{2}{13}$

4. $\dfrac{4}{13}$

Answer

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

There are 4 kings and 4 jacks i.e. total 8 kings and jacks.

∴ The number of favourable outcomes to event 'being a king or jack' is = 8.

∴ P(a king or jack) = $\dfrac{8}{52} = \dfrac{2}{13}.$

Hence, Option 3 is the correct option.

The probability that a non-leap year selected at random has 53 Sundays is

1. $\dfrac{1}{365}$

2. $\dfrac{2}{365}$

3. $\dfrac{2}{7}$

4. $\dfrac{1}{7}$

Answer

Number of days in a non-leap year = 365 i.e. 52 weeks and 1 day.

In order to have 53 sundays, the 1 day should be a sunday.

Let A be the event of the day to be a Sunday,

∴ P(A) = $\dfrac{1}{7}$.

Hence, Option 4 is the correct option.

A bag contains 3 red balls, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is

1. $\dfrac{1}{5}$

2. $\dfrac{1}{3}$

3. $\dfrac{7}{15}$

4. $\dfrac{8}{15}$

Answer

Probability that a ball drawn from the bag at random will be neither red nor black is equal to probability of picking a white ball.

No. of white balls = 5, total balls = 3 + 5 + 7 = 15.

∴ P(neither red nor black ball) = $\dfrac{5}{15} = \dfrac{1}{3}$.

Hence, Option 2 is the correct option.

A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is

1. $\dfrac{4}{9}$

2. $\dfrac{5}{9}$

3. 0

4. 1

Answer

Total no. of balls = 4 + 5 = 9.

Let A be the event of drawing a red or green ball. As, the total no. of red and green balls = 9, so the number of favourable outcomes to event A = 9.

∴ P(A) = $\dfrac{9}{9}$ = 1.

Hence, Option 4 is the correct option.

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

1. $\dfrac{1}{5}$

2. $\dfrac{3}{5}$

3. $\dfrac{4}{5}$

4. $\dfrac{1}{3}$

Answer

Sample space = {1, 2, 3, ......., 40}, with 40 equally likely outcomes.

Let A be the event that selected ticket has a number which is a multiple of 5.

∴ A = {5, 10, 15, 20, 25, 30, 35, 40}.

Hence, the number of favourable outcomes to event A = 8.

∴ P(A) = $\dfrac{8}{40} = \dfrac{1}{5}$.

Hence, Option 1 is the correct option.

If a number is randomly chosen from the numbers 1, 2, 3, 4, ........, 25, then probability of the number to be prime is

1. $\dfrac{7}{25}$

2. $\dfrac{9}{25}$

3. $\dfrac{11}{25}$

4. $\dfrac{13}{25}$

Answer

Sample space = {1, 2, 3, ......., 25}, with 25 equally likely outcomes.

Let A be the event that chosen number is prime.

∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23}.

Hence, the number of favourable outcomes to event A = 9.

∴ P(A) = $\dfrac{9}{25}$.

Hence, Option 2 is the correct option.

A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is

1. $\dfrac{1}{10}$

2. $\dfrac{9}{100}$

3. $\dfrac{1}{9}$

4. $\dfrac{3}{100}$

Answer

Sample space = {1, 2, 3, ...., 90}, which has 90 equally likely outcomes.

Let A be the event of drawing a perfect square card.

∴ A = {1, 4, 9, 16, 25, 36, 49, 64, 81}.

Hence, the number of favourable outcomes to event A = 9.

∴ P(A) = $\dfrac{9}{90} = \dfrac{1}{10}$.

Hence, Option 1 is the correct option.

If a (fair) coin is tossed twice, then the probability of getting two heads is

1. $\dfrac{1}{4}$

2. $\dfrac{1}{2}$

3. $\dfrac{3}{4}$

4. 0

Answer

A coin is tossed twice,

Sample space = {HH, HT, TH, TT}.

Let A be the event of getting two heads,

∴ A = {HH}.

∴ P(A) = $\dfrac{1}{4}$.

Hence, Option 1 is the correct option.

If two coins are tossed simultaneously, then the probability of getting atleast one head is

1. $\dfrac{1}{4}$

2. $\dfrac{1}{2}$

3. $\dfrac{3}{4}$

4. 1

Answer

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

Let A be the event of getting atleast one head,

∴ A = {HH, HT, TH}.

Hence, the no. of favourable outcomes to event A = 3.

∴ P(A) = $\dfrac{3}{4}$.

Hence, Option 3 is the correct option.

Lakshmi tosses two coins simultaneously. The probability that she gets atmost one head is

1. 1

2. $\dfrac{3}{4}$

3. $\dfrac{1}{2}$

4. $\dfrac{1}{7}$

Answer

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

Let A be the event of getting atmost one head,

∴ A = {TT, HT, TH}.

Hence, the no. of favourable outcomes to event A = 3.

∴ P(A) = $\dfrac{3}{4}$.

Hence, Option 2 is the correct option.

The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is

1. 7

2. 14

3. 21

4. 28

Answer

Let the no. of bad eggs be x.

$\therefore \text{ P(getting a bad egg)} = \dfrac{x}{400} \\[1em] \Rightarrow 0.035 = \dfrac{x}{400} \\[1em] \Rightarrow x = 0.035 \times 400 \\[1em] \Rightarrow x = 14.$

Hence, Option 2 is the correct option.

Event A : The sun will rise from east tomorrow.

Event B : It will rain on Monday.

Event C : February month has 29 days in a leap year.

Which of the above event(s) has probability equal to 1 ?

1. all events A, B and C

2. both events A and B

3. both events B and C

4. both events A and C

Answer

The sun always rises from east and also in a leap year February has 29 days.

∴ Event A and Event C has probability equal to 1.

Hence, Option 4 is the correct option.

A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a black marble is :

1. 0

2. $\dfrac{1}{5}$

3. $\dfrac{2}{5}$

4. $\dfrac{3}{5}$

Answer

Since, there is no black marble in the bag.

∴ Probability of drawing a black marble = 0.

Hence, Option 1 is the correct option.