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

Answer: (D)

$\frac{2}{5}$

Answer 2

Let $E \rightarrow$ Event of head showing up $E_{1} \rightarrow$ Event of biased coin chosen $E_{2} \rightarrow$ 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?? theorem $P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)+P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)}$ $=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{3}{4}}=\frac{2}{5}$