Out of 40 consecutive natural numbers,two are | vedclass

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(B) In a set of $40$ consecutive natural numbers,there are exactly $20$ odd numbers and $20$ even numbers.For the sum of two chosen numbers to be odd,

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$\frac{20}{39}$

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(B) In a set of $40$ consecutive natural numbers,there are exactly $20$ odd numbers and $20$ even numbers.For the sum of two chosen numbers to be odd,one number must be odd and the other must be even.The total number of ways to choose $2$ numbers out of $40$ is given by ${}^{40}C_2 = \frac{40 	imes 39}{2} = 780$.The number of ways to choose one odd number and one even number is ${}^{20}C_1 	imes {}^{20}C_1 = 20 	imes 20 = 400$.Therefore,the required probability is $P = \frac{400}{780} = \frac{40}{78} = \frac{20}{39}$.

Out of $40$ consecutive natural numbers,two are chosen at random. The probability that the sum of the numbers is odd is:

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