ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS

More Exercises

Choose the correct answer from the given four options (1 to 28):

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%
Solution:
– 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) \(1- \frac { 1 }{ p } \)
Solution:
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) \(\\ \frac { 2 }{ 5 } \)
Solution:
One digit prime numbers are 2, 3, 5, 7 = 4
Probability of an even prime number (i.e , 2) = \(\\ \frac { 1 }{ 4 } \) (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) \(\\ \frac { 1 }{ 5 } \)
Solution:
Vowels of English alphabet are a, e, i, o, u = 4
One letter is selected at random.
The probability of selecting ’e’ = \(\\ \frac { 1 }{ 5 } \) (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
Solution:
A die is thrown
Total number of events = 6
Odd number less than 3 is 1 = 1
Probability = \(\\ \frac { 1 }{ 6 } \) (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) \(\\ \frac { 1 }{ 2 } \)
Solution:
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 = \(\\ \frac { 1 }{ 6 } \) (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) \(\\ \frac { 2 }{ 3 } \)
(d) 0
Solution:
A fair die is thrown once
Total number of outcomes = 6
Composite numbers are 4, 6 = 2
Probability = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \) (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) \(\\ \frac { 2 }{ 3 } \)
Solution:
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 = \(\\ \frac { 4 }{ 6 } \) = \(\\ \frac { 2 }{ 3 } \) (d)

Question 9.
Rashmi has a die whose six faces show the letters as given below :

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) \(\\ \frac { 1 }{ 6 } \)
Solution:
A die having 6 faces bearing letters A, B, C, D, A, C
Total number of outcomes = 4
Probability of getting C = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \) (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) \(\\ \frac { 21 }{ 26 } \)
Solution:
Total number of English alphabets = 26
Letter of Delhi = D, E, L, H, I. = 5
Probability = \(\\ \frac { 5 }{ 26 } \) (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
Solution:
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
Solution:
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) \(\\ \frac { 1 }{ 13 } \)
Solution:
Total number of cards = 52
Number of aces = 4
Probability of card being an ace = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \) (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) \(\\ \frac { 9 }{ 13 } \)
Solution:
Total number of cards = 52
No. of face cards = 3 × 4 = 12
.’. Probability of face card = \(\\ \frac { 12 }{ 52 } \) = \(\\ \frac { 3 }{ 13 } \) (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) \(\\ \frac { 1 }{ 2 } \)
Solution:
Total number of card = 52
No. of red face card = 3 × 2 = 6
.’. Probability = \(\\ \frac { 6 }{ 52 } \) = \(\\ \frac { 3 }{ 26 } \) (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) \(\\ \frac { 4 }{ 13 } \)
Solution:
Total number of cards 52
Number of a king or a jack = 4 + 4 = 8
.’. Probability = \(\\ \frac { 8 }{ 52 } \) = \(\\ \frac { 2 }{ 13 } \) (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) \(\\ \frac { 1 }{ 7 } \)
Solution:
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 = \(\\ \frac { 1 }{ 7 } \) (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) \(\\ \frac { 8 }{ 1 } \)
Solution:
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 = \(\\ \frac { 5 }{ 15 } \) = \(\\ \frac { 1 }{ 3 } \) (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
Solution:
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 (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) \(\\ \frac { 1 }{ 4 } \)
Solution:
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 = \(\\ \frac { 5+4 }{ 12 } \) = \(\\ \frac { 9 }{ 12 } \) = \(\\ \frac { 3 }{ 4 } \) (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) \(\\ \frac { 1 }{ 3 } \)
Solution:
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 = \(\\ \frac { 8 }{ 40 } \) = \(\\ \frac { 1 }{ 5 } \) (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) \(\\ \frac { 13 }{ 25 } \)
Solution:
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 = \(\\ \frac { 9 }{ 25 } \) (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) \(\\ \frac { 1 }{ 100 } \)
Solution:
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 = \(\\ \frac { 9 }{ 90 } \) = \(\\ \frac { 1 }{ 10 } \) (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
Solution:
A coin is tossed twice
Number of outcomes = 2 x 2 = 4
Probability of getting two heads (HH = 1) = \(\\ \frac { 1 }{ 4 } \) (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
Solution:
Two coins are tossed
Total outcomes = 2 × 2 = 4
Probability of getting atleast one head (HT,TH,H,H) = \(\\ \frac { 3 }{ 4 } \) (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) \(\\ \frac { 1 }{ 7 } \)
Solution:
Two coins are tossed
Total number of outcomes = 2 × 2 = 4
Probability of getting atleast one head = (HT, TH, RH = 3) = \(\\ \frac { 3 }{ 4 } \) (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
Solution:
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 (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
Solution:
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 (c)

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