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(B) The total number of ways to select $2$ numbers from $40$ is $^{40}C_2 = \frac{40 	imes 39}{2} = 780$. For the sum of two numbers to be odd,one mu

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$\frac{20}{39}$

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(B) The total number of ways to select $2$ numbers from $40$ is $^{40}C_2 = \frac{40 	imes 39}{2} = 780$. For the sum of two numbers to be odd,one must be even and the other must be odd. There are $20$ even numbers and $20$ odd numbers in the first $40$ natural numbers. The number of ways to select one even and one odd number is $^{20}C_1 	imes ^{20}C_1 = 20 	imes 20 = 400$. The required probability is $P = \frac{400}{780} = \frac{40}{78} = \frac{20}{39}$.

Two numbers are selected at random from the first $40$ natural numbers. What is the probability that their sum is odd?

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