Statistics - Probabilities
- Option : B
- Explanation : 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
- Option : C
- Explanation : 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
=
34. In an experiment a coin is tossed 4 times. What is the size of the sample space?
- Option : C
- Explanation : Sample space, S = (HHHH, HHHT, HHTH, HHTT, ... , TTTT)
Clearly, |S| = 24 = 16
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