Bag A contains 2 unbiased and 3 biased coins whereas Bag B contains 3 unbiased and 2 biased coins. A bag is selected at random and 2 coins are taken out simultaneously. The probability, that both coins are unbiased is:

$\frac{1}{5}$

The correct answer is Option (1) → $\frac{1}{5}$

Let $U$ = unbiased coin, $B$ = biased coin.

Bag A: $2U, 3B$

Bag B: $3U, 2B$

Probability of choosing Bag A = $\frac{1}{2}$

Probability of choosing Bag B = $\frac{1}{2}$

From Bag A:

$P(\text{both unbiased}) = \frac{\frac{2 \times 1}{2!}}{\frac{5 \times 4}{2!}} = \frac{1}{10}$

From Bag B:

$P(\text{both unbiased}) = \frac{\frac{3 \times 2}{2!}}{\frac{5 \times 4}{2!}} = \frac{3}{10}$

Total probability:

$P = \frac{1}{2} \cdot \frac{1}{10} + \frac{1}{2} \cdot \frac{3}{10} = \frac{1}{20} + \frac{3}{20} = \frac{4}{20} = \frac{1}{5}$