Quantitative Aptitude MCQ - Probability
The sum 6 can be obtained as follows:
(1,5), (2, 4), (3, 3), (4, 2), (5, 1), i.e. in 5 ways.
Probability of A's throwing 6 with 2 dice = 5/36
Probability of A's not throwing 6 = 31/36
Similarly, probability of B's throwing 7 = 6/36, i.e. 1/6
Probability of B's not throwing 7 = 5/6
Now A can win, ifhe throws 6 in the first, third,
fifth, seventh etc. throws.
Chance of A's winning
(1,5), (2, 4), (3, 3), (4, 2), (5, 1), i.e. in 5 ways.
Probability of A's throwing 6 with 2 dice = 5/36
Probability of A's not throwing 6 = 31/36
Similarly, probability of B's throwing 7 = 6/36, i.e. 1/6
Probability of B's not throwing 7 = 5/6
Now A can win, ifhe throws 6 in the first, third,
fifth, seventh etc. throws.
Chance of A's winning
Let the event of drawing a faulty item from any of the machines be A, and the event that an item drawn at random was produced by Mi be Bi' To P (Bi/A) proceed as follows:
The highest output being from M3 required probability
=Â 
Sample space, S = (HHHH, HHHT, HHTH, HHTT, ... , TTTT)
Clearly,        |S| = 24 = 16
Clearly,        |S| = 24 = 16
