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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 the value of 81P is_____

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answer is 36.

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Detailed Solution

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Let $E_{1}, E_{2}, E_{3}$ be the events that two headed coin, biased coin, unbiased coin, is chosen respectively $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3} let E$ be the even that heads show up
Question Image by bayes theorem $P (\frac{E}{E_{1}}) = \frac{P (E_{1}) P (E / E_{1})}{P (E)}$
$= \frac{(\frac{1}{3}) (1)}{(\frac{1}{3}) (1) + \frac{1}{3} (\frac{3}{4}) + \frac{1}{3} (\frac{1}{2})} = \frac{4}{9}$

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