In class XII, suppose 5% of boys and 0.25% of girls are physically fit for a game. A fit student is selected at random from this class having same number of boys and girls. If the probability that the selected student is a girl is $\frac{m}{n}; gcd(m,n) = 1$, then $m+n$ is equal to

22

The correct answer is Option (1) → 22

Let total number of boys = total number of girls = 100 (assume equal number)

Number of fit boys = 5% of 100 = 5

Number of fit girls = 0.25% of 100 = 0.25

Total number of fit students = 5 + 0.25 = 5.25

Probability that selected fit student is a girl:

$= \frac{0.25}{5.25} = \frac{0.25}{21 \times \frac{1}{4}} = \frac{1}{21}$

So, required probability = $\frac{1}{21}$ → $m = 1$, $n = 84$

$m + n = 1 + 21 = {22}$