Question

There are two coins, one unbiased with probability $\frac{1}{2}$ of getting heads and the other one is biased with probability $\frac{3}{4}$ of getting heads. A coin is selected at random and tossed. It shows heads up. Then, the probability that the unbiased coin was selected is

• A. $\frac{2}{3}$
• B. $\frac{3}{5}$
• C. $\frac{1}{2}$
• D. $\frac{2}{5}$

Answer 1

The correct option is D $\frac{2}{5}$

Let $E\to$ Event of head showing up
$E_1\to$ Event of biased coin chosen
$E_2\to$ Event of unbiased coin chosen
Now, $P\left(E_2\right)=\frac{1}{2}$ and $P\left(E_1\right)=\frac{1}{2}$
Also, $P\left(\frac{E}{E_2}\right)=\frac{1}{2}$ and $P\left(\frac{E}{E_1}\right)=\frac{3}{4}$
(by conditional probability)
By Baye's theorem
$P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right).P\left(\frac{E}{E_2}\right)}{P\left(E_2\right).P\left(\frac{E}{E_2}\right)+P\left(E_1\right).P\left(\frac{E}{E_1}\right)}$
$=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{3}{4}}=\frac{2}{5}$