Probability Ques 104
A box contains $N$ coins, $m$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed, is $1 / 2$, while it is $2 / 3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?
(2002, 5M)
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Answer:
Correct Answer: 104.$\frac{9m}{8N+m} \quad$
Solution:
- Let $E$ be the event that coin tossed twice, shows head at first time and tail at second time and $F$ be the event that coin drawn is fair.
$ \begin{aligned} P(F / E) & =\frac{P(E / F) \cdot P(F)}{P(E / F) \cdot P(F)+P\left(E / F^{\prime}\right) \cdot P\left(F^{\prime}\right)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{m}{N}}{\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{m}{N}+\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{N-m}{N}} \\ & =\frac{\frac{m}{4}}{\frac{m}{4}+\frac{2(N-m)}{9}}=\frac{9 m}{8 N+m} \end{aligned} $
BETA
