1. The probability that a man will be alive for 10 more years is ¼ and the probability that his wife will
alive for 10 more years is 1/3. The probability that none of them will be alive for 10 more years, is
1) 5/12 2) 1/2 3) 7/12 4) 11/12 5) None of these
Solutions : (2)
Required probability = P (Ā) x P (B)
= ( 1 – ¼ ) x ( 1- 1/3 ) = ¾ x 2/3 = ½
2. In a lottery 10000 tickets are sold and ten prizes are awarded. What is the probability of not getting
a prize, if you buy one ticket?
1) 9/10000 2) 9/10 3) 999/1000 4) 9999/10000 5) None of these
Solution: (3)
Total lottery tickets = 10000
Total prize in the lottery = 10
Therefore probability of getting a prize = 10/10000 = 1/1000
Now, probability of not getting a prize = 1 – probability of getting a prize
= 1 – 1/1000 = 999/1000
3. Two persons A and B appear in an interview for two vacancies. If the probabilities of their
selections are ¼ and 1/6, respectively, then the probability that none of them is selected , is
1) 5/8 2) 5/12 3) 1/12 4) 1/24 5) None of these
Solutions : (1)
Required probability = = P (Ā) x P (B)
= ( 1- ¼) ( 1 - 1/6) = ¾ x 5/6 = 5/8
4. The probabilities of solving a problem by three students A, B and C are ½, 1/3 and ¼, respectively.
The probability that the problem will be solved, is
1) 1/4 2) 1/2 3) 3/4 4) 1/3 5) None of these
Solutions : (3).
First, we find the probability of not solving the problem.
P (A) x P (B) x P (C)
= ( 1 – ½) x ( 1- 1/3) x ( 1 – ¼)
= ½ x 2/3 x ¾ = ¼
Therefore required probability = 1- ¼ = ¾
5. A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15. The
chance that the first roll was four is
1) 2/5 2) 1/5 3) 1/6 4) None of these 5) Cannot be determined
Solutions : (4)
Total number of favourable outcomes n(S) = 63 = 216
Combinations of outcomes for getting sum of 15 on uppermost face = (4,5,6),
(5,4,6),(6,5,4),(5,6,4),(4,6,5),(6,4,5),(5,5,5),(6,6,3),(6,3,6),(3,6,6)
Now, outcomes on which first roll was a four, n(E) = (4,5,6),(4,6,5)
Therefore P(E) = n(E)/n(S) = 2/216 = 1/108
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