Question

There are three coins. One is two headed coin, 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?

Answer 1

Let E₁, E₂ and E₃ denote the events of choosing a two-headed coin, a biased coin and an unbiased coin, respectively.

Let A be the event that the coin shows heads.

\[\therefore P\left( E_1 \right) = \frac{1}{3}\]
\[ P\left( E_2 \right) = \frac{1}{3}\]
\[ P\left( E_3 \right) = \frac{1}{3}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = 1\]
\[P\left( A/ E_2 \right) = 75 \% = \frac{3}{4}\]

\[P\left( A/ E_2 \right) = \frac{1}{2}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}\]
\[ = \frac{4}{9}\]