(A) Total number of students = $15 + 5 = 20$. Number of ways to select $3$ students out of $20$ is $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times

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(A) Total number of students = $15 + 5 = 20$. Number of ways to select $3$ students out of $20$ is $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$. Number of ways to select $2$ boys out of $15$ is $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$. Number of ways to select $1$ girl out of $5$ is $^5C_1 = 5$. Required probability = $\frac{^{15}C_2 \times ^5C_1}{^{20}C_3} = \frac{105 \times 5}{1140} = \frac{525}{1140} = \frac{35}{76}$.

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$A$ class has $15$ boys and $5$ girls. Suppose $3$ students are selected at random from the class. The probability that there are $2$ boys and $1$ girl is

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A
$\frac{35}{76}$
B
$\frac{35}{38}$
C
$\frac{7}{76}$
D
$\frac{35}{72}$

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