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Q.

There are three coins. One is two headed coin (having head on both faces),another is biased coin that comes up heads 75% of the time and third is an unbiased coin, one of the three coins is chosen at random and tossed. If it shows heads, then the probability that it was 2 headed coin is p then $81 p =$

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Detailed Solution

$E_{1}, E_{2}, E_{3}$ be the events that two headed coin, biased coin, unbiased coin $\Rightarrow p (E_{1}) = p (E_{2}) = p (E_{3}) = \frac{1}{3}$

E be the event that head shows $p (\frac{E}{E_{1}}) = \frac{\frac{1}{3} (1)}{\frac{1}{3} (1) + \frac{1}{3} (\frac{3}{4}) + \frac{1}{3} (\frac{1}{2})} = \frac{1}{1 + \frac{3}{4} + \frac{1}{2}} = \frac{4}{4 + 3 + 2} = \frac{4}{9} = p \Rightarrow 81 p = 36$

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There are three coins. One is two headed coin (having head on both faces),another is biased coin that comes up heads 75% of the time and third is an unbiased coin, one of the three coins is chosen at random and tossed. If it shows heads, then the probability that it was 2 headed coin is p then 81p=