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 :

ABCDAC

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)

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