Probability Question 124
Question: A box contains N cons, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $ \frac{1}{2}, $ while it is $ \frac{2}{3} $ when a biased coin is tossed. A coin is drawn from the box at random and it tossed twice. Then the probability that the coin drawn is fair, is
Options:
A) $ \frac{9m}{8N+m} $
B) $ \frac{9m}{8N-m} $
C) $ \frac{9m}{8m-N} $
D) $ \frac{9m}{8m+N} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ E_1: $ Coin is fair, $ E_2: $ coin is biased, a second toss shows tail.
$ P(E_1/A)=\frac{P(A/E_1)P(E_1)}{P(A/E_1)P(E_1)+P(A/E_1)P(E_2)} $
$ =\frac{\frac{m}{N}.\frac{1}{2}.\frac{1}{2}}{\frac{m}{N}.\frac{1}{2}.\frac{1}{2}+\frac{N-m}{N}.\frac{2}{3}.\frac{1}{3}}=\frac{9m}{8N+m} $
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