There are three coins. One is a bvo headed coin (having head on both , another is a biased coin that comes up with heads 75\% of the time and third is an unbiased coin. One ofthe three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

.: Let ${E_1}$ : ‘coin chosen is two headed’,

${E_2}$ : ‘coin chosen is biased’ and

${E_3}$ : ‘coin chosen is unbiased’,

$\Rightarrow$ $P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = P\left( {{E_3}} \right) = \cfrac{1}{3}$

Let $A$ : 'tossed coin shows up a head',

then '
$P\left( {A|{E_1}} \right) = 1,P\left( {A|{E_2}} \right) = \cfrac{{75}}{{100}} = \cfrac{3}{4}$

and $P\left( {A|{E_3}} \right) = \cfrac{1}{2}$

(The coin, which has head on both sides shows

head with surety and an unbiased coin shows head with probability $\cfrac{1}{2}$ ).

Hence, the required probability is

$P(E/A) = \cfrac{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right)}}{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right) + P\left( {A|{E_2}} \right)P\left( {{E_2}} \right) + P\left( {A|{E_3}} \right)P\left( {{E_3}} \right)}}$

$= \cfrac{{1 \times \cfrac{1}{3}}}{{1 \times \cfrac{1}{3} + \cfrac{3}{4} \times \cfrac{1}{3} + \cfrac{1}{2} \times \cfrac{1}{3}}} = \cfrac{1}{{1 + \cfrac{3}{4} + \cfrac{1}{2}}} = \cfrac{1}{{\cfrac{{4 + 3 + 2}}{4}}} = \cfrac{4}{9}$