ML Aggarwal Solution Class 10 Chapter 22 Chapter 22 Probability MCQs

 MCQs

Question 1

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

(a) 0.7

(b) $\frac{2}{3}$

(c) – 1.5

(d) 15%

Sol :

– 1.5 (negative) can not be a probability as a probability is possible 0 to 1. (c)

Question 2

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

(a) p – 1

(b) p

(c) 1 – p

(d) $1-\frac{1}{p}$

Sol :

Complementary of p is 1 – p

Probability of complementary even of p is 1 – p. (c)

Question 3

Out of one digit prime numbers, one selecting an even number is

(a) $\frac{1}{2}$

(b) $\frac{1}{4}$

(c) $\frac{4}{9}$

(d) $\frac{2}{5}$

Sol :

One digit prime numbers are 2, 3, 5, 7 = 4

Probability of an even prime number (i.e , 2)

$=\frac{1}{4}$

Ans (b)

Question 4

Out of vowels, of the English alphabet, one letter is selected at random. The probability of selecting ‘e’ is

(a) $\frac{1}{26}$

(b) $\frac{5}{26}$

(c) $\frac{1}{4}$

(d) $\frac{1}{5}$

Sol :

Vowels of English alphabet are a, e, i, o, u = 4

One letter is selected at random.

The probability of selecting ’e’ $=\frac{1}{5}$

Ans (d)

Question 5

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

(a) $\frac{1}{6}$

(b) $\frac{1}{3}$

(c) $\frac{1}{2}$

(d) 0

Sol :

A die is thrown

Total number of events = 6

Odd number less than 3 is 1 = 1

Probability $=\frac{1}{6}$

Ans (a)

Question 6

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

(a) $\frac{1}{6}$

(b) $\frac{2}{3}$

(c) $\frac{1}{3}$

(d) $\frac{1}{2}$

Sol :

A fair die is thrown once

Total number of outcomes = 6

Prime numbers = 2, 3, 5 and even prime is 2

Probability of getting an even prime number 

$=\frac{1}{6}$

Ans (a)

Question 7

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

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{\dot{2}}{3}$

(d) 0

Sol :

A fair die is thrown once

Total number of outcomes = 6

Composite numbers are 4, 6 = 2

Probability$=\frac{2}{6}=\frac{1}{3}$

Ans (a)

Question 8

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

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{5}{6}$

(d) $\frac{2}{3}$

Sol :

A fair dice is thrown once.

Total number of outcomes = 6

Even numbers or a number greater than 4 = 2, 4, 5, 6 = 4

Probability$=\frac{4}{6}=\frac{2}{3}$

Ans (d)

Question 9

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

A

B

C

D

A

C

If she throws the die once, then the probability of getting C is

(a) $\frac{1}{3}$

(b) $\frac{1}{4}$

(c) $\frac{1}{5}$

(d) $\frac{1}{6}$

Sol :

A die having 6 faces bearing letters A, B, C, D, A, C

Total number of outcomes = 4

Probability of getting C 

$=\frac{2}{6}=\frac{1}{3}$

Ans (a)

Question 10

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

(a) $\frac{1}{5}$

(b) $\frac{1}{26}$

(c) $\frac{5}{26}$

(d) $\frac{21}{26}$

Sol :

Total number of English alphabets = 26

Letter of Delhi = D, E, L, H, I. = 5

Probability$=\frac{5}{26}$

Ans (c)

Question 11

A card is drawn from a well-shuffled pack of 52 playing cards. The event E is that the card drawn is not a face card. The number of outcomes favourable to the event E is

(a) 51

(b) 40

(c) 36

(d) 12

Sol :

Number of playing cards = 52

Probability of a card which is not a face card = (52 – 12) = 40

Number of possible events = 40 (b)

Question 12

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

Sol :

Total number of cards = 52

Balance 52 – 1 = 51

Number of possible events = 51 (d)

Question 13

If one card is drawn from a well-shuffled pack of 52 cards, the probability of getting an ace is

(a) $\frac{1}{52}$

(b) $\frac{4}{13}$

(c) $\frac{2}{13}$

(d) $\frac{1}{13}$

Sol :

Total number of cards = 52

Number of aces = 4

Probability of card being an ace

$=\frac{4}{52}=\frac{1}{13}$

Ans (d)

Question 14

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

(a) $\frac{3}{13}$

(b) $\frac{4}{13}$

(c) $\frac{6}{13}$

(d) $\frac{9}{13}$

Sol :

Total number of cards = 52

No. of face cards = 3 × 4 = 12

∴ Probability of face card $=\frac{12}{52}=\frac{3}{13}$

Ans (a)

Question 15

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

(a) $\frac{3}{26}$

(b) $\frac{3}{13}$

(c) $\frac{2}{13}$

(d) $\frac{1}{2}$

Sol :

Total number of card = 52

No. of red face card = 3 × 2 = 6

∴ Probability $=\frac{6}{52}=\frac{3}{26}$

Ans (a)

Question 16

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

(a) $\frac{1}{26}$

(b) $\frac{1}{13}$

(c) $\frac{2}{13}$

(d) $\frac{4}{13}$

Sol :

Total number of cards 52

Number of a king or a jack = 4 + 4 = 8

∴ Probability $=\frac{8}{52}=\frac{2}{13}$

Ans (c)

Question 17

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

(a) $\frac{1}{365}$

(b) $\frac{2}{365}$

(c) $\frac{2}{7}$

(d) $\frac{1}{7}$

Sol :

Number of a non-leap year 365

Number of Sundays = 53

In a leap year, there are 52 weeks or 364 days

One days is left

Now we have to find the probability of a Sunday out of remaining 1 day

∴ Probability$=\frac{1}{7}$ 

Ans (d)

Question 18

A bag contains 3 red balk, 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

(a) $\frac{1}{5}$

(b) $\frac{1}{3}$

(c) $\frac{7}{15}$

(d) $\frac{8}{1}$

Sol :

In a bag, there are

3 red balls + 5 white balls + 7 black balls

Total number of balls = 15

One ball is drawn at random which is neither

red not black

Number of outcomes = 5

Probability$=\frac{5}{15}=\frac{1}{3}$

Ans (b)

Question 19

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

(a) $\frac{4}{9}$

(b) $\frac{5}{9}$

(c) 0

(d) 1

Sol :

In a bag, there are

4 red balls + 5 green balls

Total 4 + 5 = 9

One ball is drawn at random

Probability of either a red or a green ball

$=\frac{9}{9}=1$

Ans (d)

Question 20

A bag contains 5 red, 4 white and 3 black balls. If a. ball is drawn from the bag at random, then the probability of the ball being not black is

(a) $\frac{5}{12}$

(b) $\frac{1}{3}$

(c) $\frac{3}{4}$

(d) $\frac{1}{4}$

Sol :

In a bag, there are

5 red + 4 white + 3 black balls = 12

One ball is drawn at random

Probability of a ball not black$=\frac{5+4}{12}=\frac{9}{12}=\frac{3}{4}$

Ans (c)

Question 21

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

(a) $\frac{1}{5}$

(b) $\frac{3}{5}$

(c) $\frac{4}{5}$

(d) $\frac{1}{3}$

Sol :

There are t to 40 = 40 tickets in a bag

No. of tickets which is multiple of 5 = 8

(5, 10, 15, 20, 25, 30, 35, 40)

Probability$=\frac{8}{40}=\frac{1}{5}$

Ans (a)

Question 22

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

(a) $\frac{7}{25}$

(b) $\frac{9}{25}$

(c) $\frac{11}{25}$

(d) $\frac{13}{25}$

Sol :

There are 25 number bearing numbers 1, 2, 3,…,25

Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9

Probability being a prime number$=\frac{9}{25}$

Ans (b)

Question 23

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

(a) $\frac{1}{10}$

(b) $\frac{9}{100}$

(c) $\frac{1}{9}$

(d) $\frac{1}{100}$

Sol :

In a box, there are

90 cards bearing numbers 1 to 90

Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9

Probability of being a perfect square$=\frac{9}{90}=\frac{1}{10}$

Ans (a)

Question 24

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

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{3}{4}$

(d) 0

Sol :

A coin is tossed twice

Number of outcomes = 2 x 2 = 4

Probability of getting two heads (HH = 1)

$=\frac{1}{4}$

Ans (a)

Question 25

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

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{3}{4}$

(d) 1

Sol :

Two coins are tossed

Total outcomes = 2 × 2 = 4

Probability of getting atleast one head (HT,TH,H,H) 

$=\frac{3}{4}$

Ans (c)

Question 26

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

(a) 1

(b) $\frac{3}{4}$

(c) $\frac{1}{2}$

(d) $\frac{1}{7}$

Sol :

Two coins are tossed

Total number of outcomes = 2 × 2 = 4

Probability of getting atleast one head = (HT, TH, RH = 3) 

$=\frac{3}{4}$

Ans (b)

Question 27

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

(a) 7

(b) 14

(c) 21

(d) 28

Sol :

Total number of eggs 400

Probability of getting a bad egg = 0.035

Number of bad eggs = 0.035 of 400 

$=400 \times \frac{35}{1000}=14$

Ans (b)

Question 28

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets she has bought?

(a) 40

(b) 240

(c) 480

(d) 750

Sol :

For a girl,

Winning a first prize = 0.08

Number of total tickets = 6000

Number of tickets she bought = 0.08 of 6000 

$=6000 \times \frac{8}{100}=480$

Ans (c)