# Probability Questions and Answers updated daily – Aptitude

Probability Questions: Solved 93 Probability Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Probability Questions

81. How many 4 digit even number can formed by using the digits 1,3,7 and 8 only once ?

Correct Ans:6
Explanation:
Solution is :
We can form 24 numbers with digits 1,3,7 and 8 using only once.
i.e., 4P4 = 4! = 4 * 3 * 2 * 1 = 24.

To be an even number, the units digit has to be an even number, in this case only 8 should come in the unit's place . The remaining places, 3 positions can be filled in 3! , that is 6 ways
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82. In a class there are some boys and 20 girls. The probability of selecting a girl is twice the probability of selecting a boy. Find the number of boys

Correct Ans:10
Explanation:
Solution is :
Let the number of boys be x.
Given there are 20 girls
Total number of people are (b + g) = x+20
Probability of selecting a girl, P(G) = 20 / (x+20)
Probability of selecting a boy, P(B) = x / (x+20)
Given P(G) = 2 x P(B)
=> 20 / (x+20) = 2x / (x+20)
=> 20 = 2x
=> x = 10
There are totally 10 boys
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83. Two coins are tossed. What is the probability of getting exactly one head ?

Correct Ans:1/2
Explanation:
Solution is :
Let H denote head and T denote tail.

When two coins are tossed, the possible cases are
S = {HH, HT, TH , TT}
=> n(S) = 4

Let E be the event which denotes getting exactly one head
E = {HT , TH}
=> n(E) = 2

Probability of getting exactly one head = P(E) = n(E) / n(S) = 2/4 = 1/2
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84. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers appear on the top of the dice is 7?

Correct Ans:1/6
Explanation:
Solution is :
Sample Space, S = { (1,1) , (1,2) , (1,3) , ... (6,6) }
n(S) = 6 x 6 = 36 possible cases.
E = {(x,y) / such that x+y=7}
E = {(1,6) , (2,5) , (3,4) , (4,3) , (5,2) , (6,1)} => n(E)=6

Probability, P(E) = n(E) / n(S)
= 6 / 36 = 1/6
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85. A box contains 4 black balls, 3 red balls and 5 green balls. Two balls are drawn at random. What is the probability that both the balls are of same color?

Correct Ans:19/66
Explanation:
Solution is
Given

Black ball=4 ; red ball=3 ; green ball=5
Total balls = 4 + 3 + 5 = 12
2 balls drawn from 12.
Number of possible ways = 12C2 = (12x11) / 2 = 66 ways
Number of possible ways that both balls are of same colour =>4C2 + 3C2 + 5C2
= (4x3) /2 + (3x2) /2 +(5x4) /2
= 6 + 3 + 10 = 19
Probability that both balls are of same colour = 19 / 66
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86. In a class there are some male and 30 female candidates. The probability of selecting a female is thrice the probability of selecting a male. Find the number of males.

Correct Ans:10
Explanation:
Solution is
Given

Let the number of males be x.
Given there are 30 females
Total number of people are x+30
Probability of selecting female, P(F) = 30 / (x+30)
Probability of selecting male, P(M) = x / (x+30)
Given P(F) = 3 x P(M)
=> 30 / (x+30) = 3x / (x+30) => 30 = 3x => x = 10
There are totally 10 males
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87. A box contains 4 black balls, 3 red balls and 4 green balls. Two balls are drawn at random. What is the probability that both the balls are of same color?

Correct Ans: 13/55
Explanation:
Total balls = 4+3+4=11 2 balls drawn from 11. Numbebr of possible ways = 11C2 = 11x10/2 = 55 ways Both are same color = 4C2+3C2 + 4C2 =6 + 3 + 4 = 13 Probability = 13/55
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88. Two coins are tossed. What is the probability of getting exactly one tail ?

Correct Ans:1/2
Explanation:
Let H denote head and T denote tail. When two coins are tossed, the possible cases are S = {HH, HT, TH , TT} Let E be the event which denotes getting exactly one tail E = {HT , TH} => n(E) = 2 P(E) = n(E) / n(S) = 2/4 = 1/2
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89. Two coins are tossed. What is the probability of getting atleast one tail ?

Correct Ans:3/4
Explanation:
Let H denote head and T denote tail. When two coins are tossed, the possible cases are S = {HH, HT, TH , TT} Let E be the event which denotes getting atleast one tail E = {HT , TH, TT} => n(E) = 3 P(E) = n(E) / n(S) = 3/4
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90. Two coins are tossed. What is the probability of getting atleast one tail ?

Correct Ans:3/4
Explanation:
Let H denote head and T denote tail. When two coins are tossed, the possible cases are S = {HH, HT, TH , TT} Let E be the event which denotes getting atleast one tail E = {HT , TH, TT} => n(E) = 3 P(E) = n(E) / n(S) = 3/4
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91. In a class there are some boys and 20 girls. The probability of selecting a girl is twice the probaility of selecting a boy. Find the number of boys

Correct Ans:10
Explanation:
Let the number of boys be x. Given there are 20 girls Total number of people are x+20 Probaility of selecting girl, P(G) = 20 / (x+20) Probaility of selecting boy, P(B) = x / (x+20) Given P(G) = 2 x P(B) => 20 / (x+20) = 2x / (x+20) => 20 = 2x => x = 10 There are totally 10 boys.
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92. In a class there are some boys and 20 girls. The probability of selecting a girl is twice the probability of selecting a boy. Find the number of boys

Correct Ans:10
Explanation:
Let the number of boys be x. Given there are 20 girls Total number of people are x+20 Probaility of selecting girl, P(G) = 20 / (x+20) Probaility of selecting boy, P(B) = x / (x+20) Given P(G) = 2 x P(B) => 20 / (x+20) = 2x / (x+20)=> 20 = 2x => x = 10 There are totally 10 boys
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93. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers appear on the top of the dice is 7.

Correct Ans:????? 1/4
Explanation:
Sample Space, S = { (1,1) , (1,2) , (1,3) , ... (6,6) } n(S) = 6 x 6 = 36 possible cases. E = {(x,y) / such that x+y=7} E = {(1,6) , (2,5) , (3,4) , (4,3) , (5,2) , (6,1)} => n(E)=6 P(E) = n(E)/n(S) = 6 / 36 = 1/4
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