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A box contains 5 fair coins and 5 biased coins. Each biased coin has a probability of a head 4/5. A coin is drawn at random from the box and tossed. Then a second coin is drawn at random from the box (without replacing the first one.) Given that the first coin has shown head, the conditional probability that the second coin is fair, is 

a) 20/39

b) 20/37

c) 1/2

d) 7/13

1 Answer

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4 votes

Required probability = P(second coin = fair / first coin shows head)

By conditional probability theorem, P(A / B) = $\frac{P(A \cap B)}{P(B)}$

$\therefore$ P(second coin = fair / first coin shows head) = $\frac{P(second coin = fair \cap first coin shows head)}{P(first coin shows head)}$

Now consider, 

P(first coin shows head) = probability of getting a head on a fair coin + probability of getting a head on a biased coin

                                       = $\frac{5}{10} * \frac{1}{2} + \frac{5}{10} * \frac{4}{5}$

                                       = $\frac{13}{20}$

Now, P(second coin = fair $\cap$ first coin shows head)

If we observe closely, we can conclude that these probabilities are independent probabilities and for two independent events A and B, P(A $\cap$ B) = P(A) * P(B)

$\therefore$ P(second coin = fair $\cap$ first coin shows head)

= P(first coin shows head) * P(second coin = fair)

= $\frac{5}{10} * \frac{1}{2} * \frac{4}{9}$ +$\frac{5}{10} * \frac{4}{5} * \frac{5}{9}$

= $\frac{1}{3}$

$\therefore$ Required probability = $\frac{\frac{1}{3}}{\frac{13}{20}}$ = $\frac{20}{39}$

So, a) is correct.

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