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A box contains $5$ fair and $5$ biased coins. Each biased coin has a probability of head $\frac{4}{5}$. A coin is drawn at random from the box and tossed. Then the 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

  1. $\frac{20}{39}\\$
  2. $\frac{20}{37}\\$
  3. $\frac{1}{2}\\$
  4. $\frac{7}{13}$
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prob of head from fair coin= (5*1/2)/((5*1/2)+(5*4/5))=5/13

prob of head from biased coin= (5*4/5)/((5*1/2)+(5*4/5))=8/13

now probability of second coin to be fair is given as =  (head from baised coin*probability of fair coin if biased coin was selected at first )+(head from fair coin*probability of fair coin if fair coin was selected at first)= 8/13*5/9 + 5/13*4/9 = 20/39.

Ans-A

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P(A):The probability of getting head is $\frac{5}{10}*\frac{1}{2}+\frac{5}{10}*\frac{4}{5}=\frac{13}{20}$

Probability that second coin is= first fair * second fair + first unfair * second fair=

P(B):Probability that second coin is fair= $\frac{1}{4}*\frac{4}{9}+\frac{2}{5}*\frac{5}{9}=\frac{1}{3}$

conditional probability = $\frac{P(B)}{P(A)}=\frac{\frac{1}{3}}{\frac{13}{20}}=\frac{20}{39}$

Option A

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