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 an unbiased coin. One of the three coins is chosen at random and tossed, it shows head, what is the probability that it is was the two headed coin?

Answer 1

Let $E_1$ : the event that the coin chosed is two headed, $E_2$: the event that the coin chosen is biased and $E_3$ : the event that the coin chosen is unbiased

$\Rightarrow E_1,E_2,E_3$are mutually exclusive and exhaustive events. Moreover,

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

Let E: tosses coin shows up a head,

$thereforeP\frac{E}{E_1}$ = P(coin showing heads, given that it is a two headed coin)=1

$P\frac{E}{E_2}$ = P(coin showing heads, given that it is a biased coin)

75%= $\frac{75}{100}=\frac{3}{4}$

$P\frac{E}{E_3}$ =P(coin showing heads, given that it is an unbiased coin)=$\frac{1}{2}$

The probability that the coiin is two headed, given that it shows head, is given by $P\frac{E_1}{E}$

By using Baye's theorem, we obtain

$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)+P\left(\frac{E}{E_3}\right)P\left(E_3\right)}$

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