Question

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 times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

Answer 1

$\text{Let events}{E}_1,E_2\text{and}{E}_3\text{denotes as}$

$E_1$: $\text{coin chosen is two headed,}$

$E_2$ : $\text{coin chosed is biased with heads}75\%$

$E_3$ : $\text{coin chosen is also biased with heads}60\%$

$\text{then}{E}_1,E_2,E_3\text{are mutually exclusive and exhaustive.}$

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

$\text{Let A denotes : Tossed coin shows up a head}$

$\therefore P\left(A/E_1\right)=1,P\left(A/E_2\right)=\frac{75}{100}=\frac{3}{4},P\left(A/E_3\right)=\frac{60}{100}=\frac{3}{5}$

$\text{Required probability=}$$P\left(E_1/A\right)$

= $\frac{P\left(A/E_1\right)P\left(E_1\right)}{P\left(A/E_1\right)P\left(E_1\right)+P\left(A/E_2\right)P\left(E_2\right)+P\left(A/E_3\right).P\left(E_3\right)}$

(By Baye's Theorm)

= $\frac{1\times \frac{1}{3}}{1\times \frac{1}{3}+\frac{3}{4}\times \frac{1}{3}+\frac{3}{5}\times \frac{1}{3}}=\frac{1}{1+\frac{3}{4}+\frac{3}{5}}$

= $\frac{1}{\frac{20+15+12}{20}}=\frac{20}{47}$