There are three coins. One is two headed coin (having head on both faces),

Solution:

Let E1, E2, and E3 be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.

$\therefore P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=\frac{1}{3}$

Let A be the event that the coin shows heads.

A two-headed coin will always show heads.

$\therefore \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}($ coin showing heads, given that it is a two-headed coin $)=1$

Probability of heads coming up, given that it is a biased coin= 75%

$\therefore \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P}($ coin showing heads, given that it is a biased coin $)=\frac{75}{100}=\frac{3}{4}$

Since the third coin is unbiased, the probability that it shows heads is always $\frac{1}{2}$.

$\therefore \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{3}\right)=\mathrm{P}($ coin showing heads, given that it is an unbiased coin $)=\frac{1}{2}$

The probability that the coin is two-headed, given that it shows heads, is given by

$P\left(E_{1} \mid A\right)$

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)+P\left(E_{3}\right) \cdot P\left(A \mid E_{3}\right)}$

$=\frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1+\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{3} \cdot \frac{1}{2}}$

$=\frac{\frac{1}{3}}{\frac{1}{3}\left(1+\frac{3}{4}+\frac{1}{2}\right)}$

$=\frac{1}{\frac{9}{4}}$

$=\frac{4}{9}$