Problems on Ages
Problems on Ages - 1 Problems on Ages - 2 Problems on Ages - 3 Problems on Ages - 4Problems on Average
Problems on Average - 1 Problems on Average - 2 Problems on Average - 3Odd Man Out Series
Odd Man Out Series - 1 Odd Man Out Series - 2 Odd Man Out Series - 3 Odd Man Out Series - 4Problems on Percentage
Problems on Percentage - 1 Problems on Percentage - 2 Problems on Percentage - 3 Problems on Percentage - 4 Problems on Percentage - 5Permutation and Combination
Permutation and Combination - 1 Permutation and Combination - 2 Permutation and Combination - 3Problems on Probability
Problems on Probability - 1 Problems on Probability - 2 Problems on Probability - 3 Problems on Probability - 4Problems on Profit Loss
Profit and Loss - 1 Profit and Loss - 2 Profit and Loss - 3 Profit and Loss - 4 Profit and Loss - 5 Profit and Loss - 6Problems on Test Divisibility
Test Divisibility - 1 Test Divisibility - 2 Test Divisibility - 3Time and Work
Time and Work - 1 Time and Work - 2 Time and Work - 3The ones who are crazy enough to think they can change the world are the ones who do.- Steve Jobs
Problems on Probability
Probability - Aptitude
1.In a throw of a coin, find the probability of getting a head.
x
Option: D
Explanation
S = {H, T}
E = {H}
P(E) = n(E)/n(S) = 1/2
2.Two unbiased coins are tossed. What is the probability of getting at most one head?
x
Option: A
Explanation
S = {HH, HT, TH, TT}
E = event of getting at most head
E = {HH, HT, TH}
P(E) = n(E)/n(S) = 3/4
3.An unbiased die is tossed. Find the probability of getting a multiple of 3.
x
Option: D
Explanation
S = {1, 2, 3, 4, 5, 6}
E = event of getting multiple of 3
E = {3, 6}
P(E) = n(E)/n(S) = 2/6 = 1/3
4.In a simultaneous throw of a pair of dice, find the probability of getting a total more than 7.
x
Option: A
Explanation
n(S) = {6 * 6} = 36
E = event of getting a total more than 7
E = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
P(E) = n(E)/n(S) = 15/36 = 5/12
5.A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of the same colour.
x
Option: A
Explanation
n(S) = number of ways of drawing 2 balls out of (6 + 4) = 10C2
n(S) = (10 * 9)/(2 * 1) = 45
E = event of getting both balls of the same colour
n(E) = number of ways of drawing (2 balls out of 6) or (2 balls out of 4)
n(E) = (6C2 + 4C2) = ((6 * 5)/(2 * 1)) + ((4 * 3)/(2 * 1)) = 15 + 6 = 21
P(E) = n(E)/n(S) = 21/45 = 7/15
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