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…

The class has equal numbers of boys and girls, with 5% of boys and 0.25% of girls physically fit. A fit student is selected at random, and we need to find the probability that the selected student is a girl.

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Let there be $n$n boys and $n$n girls in the class.

Number of fit boys:

$$\frac{5}{100} \times n$$1005​×n

$$= \frac{n}{20}$$=20n​

Number of fit girls:

$$\frac{0.25}{100} \times n$$1000.25​×n

$$= \frac{1}{400} \times n$$=4001​×n

$$= \frac{n}{400}$$=400n​

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Total number of fit students:

$$\frac{n}{20} + \frac{n}{400}$$20n​+400n​

Converting to common denominator:

$$= \frac{20n}{400} + \frac{n}{400}$$=40020n​+400n​

$$= \frac{21n}{400}$$=40021n​

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The probability that the selected fit student is a girl:

$$P(\text{girl | fit}) = \frac{\text{Number of fit girls}}{\text{Total fit students}}$$P(girl | fit)=Total fit studentsNumber of fit girls​

$$= \frac{\frac{n}{400}}{\frac{21n}{400}}$$=40021n​400n​​

$$= \frac{n}{400} \times \frac{400}{21n}$$=400n​×21n400​

$$= \frac{1}{21}$$=211​

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The probability is $\frac{1}{21}$211​ where $m = 1$m=1 and $n = 21$n=21.

Since $\gcd(1, 21) = 1$gcd(1,21)=1, the answer is:

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