Exercise: Problems on Probability

This is the aptitude questions and answers section on 'Probability' with solutions and detailed explanation. Questions on the basic definitions and theorems of probability are asked.

Problems on Probability : Question 1 :
If two dice are thrown simultaneously, then find the probability that the sum of numbers appeared on the dice is 6 or 7?
5/6
7/36
5/36
11/36
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Explanation:
The sum of numbers appeared is 6 or 7. Therefore, the required sums are 6 or 7, i.e., the required events are (1,5), (5,1), (2,4), (4,2), (3,3), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)
i.e., for 6, n(E) = 5 and for 7, n(E) = 6
Therefore, the required probability = n(E)/n(S) = 11/36.
Problems on Probability : Question 2 :
The probability that Alex will solve a problem is 1/5. What is the probability that he solves at least one problem out of 4 problems?
64/125
256/625
64/625
369/625
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Explanation:
The probability that Alex will not solve a problem = 4/5. The probability that Alex will not solve 10 problems = (4/5)4 = 256/625. Hence, the probability that Alex will solve at least one problem = 1 - 256/625 = 369/625.

Problems on Probability : Question 3 :
In a race where 12 cars are running, the chance that car X will win is 1/6, that Y will win is 1/10 and that Z will win is 1/8. Assuming that a dead heat is impossible, find the chance that one of them will win.
1/140
1/180
27/410
47/120
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Explanation:
Required probability = P(X) + P(Y) + P(Z) (all the events are mutually exclusive).
= 1/6 + 1/10 + 1/8 = 47/120
Problems on Probability : Question 4 :
Out of 17 applicants, there are  8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.
25/34
19/34
21/34
27/34
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Explanation:
The events of selection of two person is redefined as (i) first is a girl AND second is a boy OR (ii) first is boy AND second is a girl OR (iii) first is a girl and second is a girl.
So the required probability:
= (8/17 x 9/16) + (9/17 x 8/16) + (9/17 x 8/16)
= 9/34 + 9/34 + 9/34
= 27/34.
Problems on Probability : Question 5 :
What is the probability that in the rearrangements of the word AMAZED, the letter ’E’ is positioned in between the 2 ’A’s (Not necessarily flanked)?

1/6
1/4
1/3
1/5
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Explanation:
In any rearrangement of the word, consider only the positions of the letters A, A and E.
These can be as A A E, A E A or E A A.
Therefore, effectively one-third of all words will have ’E’ in between the two ’A’s.