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There are three coins. One is a two headed coin (having head on both faces), another is a 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, it shows heads, what is the probability that it was the two headed coin ?


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  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $(\;P(E_i/A))=(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)})$
Let $E_1$: event that the coin is 2 headed, $E_2$ be the event that its biased with heads 75% of the time and $E_3$ be the fair coin.
All these three events are mutually exclusive and exchaustive, and are equally likely.
$\therefore, P(E_1) = P(E_2) = P(E_3) = \large\frac{1}{3}$
Let one of the coin chosen at random be tossed up.
Let $E$ be the event of chosen coin showing head.
$P ($coin shows head given that its 2 headed coin) $= P (E|E_1) = 1$.
$P ($coin shows head given that its 75% biased for heads) $= P (E|E_2) = \large\frac{3}{4}$
$P ($coin shows head given that its a fair coin)$ = P (E|E_3) = \large\frac{1}{2}$
We need to find the probability that it is two headed coin, given that it shows head.
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)+ P(E_3)P(A|E_3)}$
Therefore $ P(E_1|E) = \large\frac{\frac{1}{3} }{\frac{1}{3} + \frac{1}{3}.\frac{3}{4} + \frac{1}{3}.\frac{1}{2}}$
$P(E_1|E) = \large\frac{1}{1 + \frac{3}{4} + \frac{1}{2}} = \large\frac{1}{\frac{4+3+2}{4}} = \frac{4}{9}$
answered Jun 19, 2013 by balaji.thirumalai
edited Jan 4 by priyanka.c

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