Solved Questions on Probability
1. The three events P, Q and R are as follows:
Event P: If an integer is chosen at random from 1 to 50, then the probability that the number is ‘divisible by 5’.
Event Q: A box contains 2 red, 3 black and 5 white balls. If a ball is drawn at random, then the probability that the ball drawn is a ‘red ball’.
Event R: English letters are arranged in a row. If a letter is chosen at random from the letters of the English alphabet, then the probability is that it is a letter of the word ‘JAGUAR’.
Which of the above events have probabilities equal to 0.2?
a) Both events P and Q
b) Both events Q and R
c) Both events R and P
d) All events P, Q and R
Answer: a) Both events P and Q
Explanation:
Event P:
Total number of outcomes = 50
Number of favourable outcomes {5, 10, 15, 20, 25, 30, 35, 40, 45, 50} = 10
Required Probability = 10/ 50 = 1/5 = 0.2
Event Q:
Number of red balls = 2
Number of black balls = 3
Number of white balls = 5
Total number of balls = 2 + 3 + 5 = 10
Probability of getting a red ball = 2/10 = 1/5 = 0.2
Event R:
Total number of English alphabets = 26
Letters of JAGUAR {A, G, J, U, R} = 5
Required Probability = 5/26 = 0.19
∴ Both events P and Q have probabilities equal to 0.2.
2. A bag contains some blue balls and 45 red balls. If the probability of drawing a blue ball is three-fifths of a red ball, then what is the number of blue balls in the bag?
a) 21
b) 23
c) 27
d) 29
Answer: c) 27
Explanation: Let the bag contain x blue balls.
Number of red balls = 45
Total number of balls in a bag = x + 45
Probability of drawing a blue ball = x⁄x + 45
Probability of drawing a red ball = 45⁄x + 45
According to the question,
Probability of drawing a blue ball is three-fifths of a red ball.
⇒ x⁄x +m 45 = 3⁄5 × 45⁄x + 45
⇒ x = 3×45⁄5
∴ x = 27
3. Cards are labelled as c, d, e,....., s, t. They are put in a box and shuffled. A student is asked to draw a card from the box. What is the probability that the card draws none of the letters of the word ‘jacket’?
a) 33
b) 53
c) 67
d) 87
Answer: c) 0.67
Explanation: S = {c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t}
Total number of cards = n(S) = 18
E = {j, a, c, k, e, t} = {a, c, e, k, j, t}
Number of the letters in the word ‘jacket’ = n(E) = 6
Number of letters not in the word ‘jacket’ = n(E′) = n(S) − n(E) = 18 − 6 = 12
Probability that the card draws none of the letters of the word ‘jacket’
= n(E′)⁄n(S) = 12⁄18 = 0.67
4. The events are as follows:
Event C: If a card is selected at random from a pack of 52 cards, then a blackface card is found.
Event D: If a dice is thrown, then an odd number more than 2 is found on the top of a dice.
Event T: If a ticket is drawn at random from a box containing tickets numbered 1 to 25, then the selected ticket has a number which is a multiple of 7.
Which of the above events have probabilities equal to 1/3?
a) Both events C and D
b) Both events D and T
c) Both events T and C
d) All events C, D and T
Answer: b) Both events D and T
Explanation:
Event C: Total number of cards = 52
Number of black face cards = 3 × 2 = 6
Required Probability = 6/52 = 3/26
Event D: A die is thrown.
Total number of events = 6
Total number of odd numbers more than 2 = 2 (i.e., 3, 5)
Required Probability = 2/6 = 1/3
Event T: There are a total 25 tickets in a bag.
Number of tickets which is multiple of 7 = 3 (7, 14 and 21)
Required Probability = 3⁄15 = 1/3
Both events D and T have probabilities equal to 1/3.
5. What is the probability when two dice are thrown simultaneously whose sum is at most 7?
a) 36/42
b) 36/49
c) 49/64
d) 49/84
Answer: d) 49/84
Explanation: The sum includes numbers that are both less than 7 and equal to 7. The sum whose at most is 7 is marked as
Total Outcomes = n(s) = 36
Favourable Outcomes = n(E) = 21
Probability when two dice are thrown simultaneously whose sum is at most 7.
P(E) = n(E)/n(s) = 21/36 = 7/12
7/12 = 49/84
