Q63 / 200 · WBJEE 2013 · Mathematics
There are two coins, one unbiased with probaility $\frac{1}{2}$ or 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 wa selected is
- A$\frac{2}{3}$
- B$\frac{3}{5}$
- C$\frac{1}{2}$
- Correct answerD$\frac{2}{5}$
Solution
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's theorem
$$
\begin{aligned}
P\left(\frac{E_{2}}{E}\right)=& \frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)+P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)} \\
&=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{3}{4}}=\frac{2}{5}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.