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 unbiasedcoin 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 $\frac{2}{5}$

Consider $2$ coins,

In the care of unbiased coin, probability of getting head $=P\left(E_1\right)=\left(\frac{1}{2}\right)$

In the care of biased coin, probability of getting head $=P\left(E_2\right)=\left(\frac{3}{4}\right)$

The probability of getting a biased or unbiased coin $=P\left(\frac{A}{E_1}\right)=P\left(\frac{A}{E_2}\right)=\frac{1}{2}$

$\Rightarrow A=$ Probability that unbiased coin is selected.

$\Rightarrow P\left(\frac{E}{E_1}\right)=\frac{P\left(E_1\right)P\left({\frac{A}{E}}_1\right)}{P\left(E_1\right)P\left({\frac{A}{E}}_1\right)+P\left(E_2\right)P\left({\frac{A}{E}}_2\right)}$

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

$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{3}{8}}$

$=\frac{\frac{1}{4}}{\frac{5}{8}}$

$=\frac{2}{5}.$

Hence, the answer is $\frac{2}{5}.$