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(B) Total number of ways to select three numbers from $50$ is $^{50}C_3 = \frac{50 	imes 49 	imes 48}{3 	imes 2 	imes 1} = 19,600$. For three numb

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$\frac{3}{98}$

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(B) Total number of ways to select three numbers from $50$ is $^{50}C_3 = \frac{50 	imes 49 	imes 48}{3 	imes 2 	imes 1} = 19,600$. For three numbers $a, b, c$ to be in arithmetic progression, $2b = a + c$. This implies $a + c$ must be even, which occurs if both $a$ and $c$ are odd or both are even. Number of odd numbers in the set is $25$ and number of even numbers is $25$. Number of ways to choose two odd numbers is $^{25}C_2 = \frac{25 	imes 24}{2} = 300$. Number of ways to choose two even numbers is $^{25}C_2 = \frac{25 	imes 24}{2} = 300$. Total favorable cases = $300 + 300 = 600$. Probability = $\frac{600}{19600} = \frac{6}{196} = \frac{3}{98}$.

If three numbers are randomly selected from the set $\{1, 2, 3, \ldots, 50\}$, then the probability that they are in arithmetic progression is

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