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

Question:

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

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

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

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

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

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

$=\frac{4}{9}$