Question

There are three coins, identical in appearance, one of which is ideal and other two biased with probabilities 1/3 and 2/3 respectively for a head. One coin is taken at random and tossed twice: if a head appears both times, what is the probability that the ideal coin was chosen ?

Answer 1

Consider the problem

Let $E_1,E_{2}and{E}_3$ be the events of choosing an ideal dice, first biased dice and second biased dice respectively. and $E$ be the event of head of appearing on both toss.

$\begin{array}{c}p\left(E_1\right)=p\left(E_2\right)=p\left(E_3\right)=\frac{1}{3}\\ p\left(\frac{E}{E_1}\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\\ p\left(\frac{E}{E_2}\right)=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}\\ p\left(\frac{E}{E_3}\right)=\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}\end{array}$

And

$\begin{array}{c}p\left(\frac{E_1}{E}\right)=\frac{p\left(E_1\right)\times p\left(\frac{E}{E_1}\right)}{p\left(E_1\right)\times p\left(\frac{E}{E_1}\right)+p\left(E_2\right)\times p\left(\frac{E}{E_2}\right)+p\left(E_3\right)\times p\left(\frac{E}{E_3}\right)}\\ =\frac{\frac{1}{3}\times \frac{1}{4}}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{3}\times \frac{1}{9}+\frac{1}{3}\times \frac{4}{9}}\\ =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{9}+\frac{4}{9}}\\ =\frac{\frac{9}{36}}{\frac{9+4+16}{36}}\\ =\frac{9}{29}\end{array}$