Out of 40 consecutive integers, two integers | vedclass

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(B) The total number of ways to select $2$ integers out of $40$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.Total outcomes

Vedclass · Accepted Answer

$\frac{20}{39}$

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(B) The total number of ways to select $2$ integers out of $40$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.Total outcomes $= {}^{40}C_{2} = \frac{40 	imes 39}{2 	imes 1} = 20 	imes 39 = 780$.For the sum of two integers to be odd, one must be even and the other must be odd.In $40$ consecutive integers, there are $20$ even integers and $20$ odd integers.The number of ways to choose one even integer from $20$ is ${}^{20}C_{1} = 20$.The number of ways to choose one odd integer from $20$ is ${}^{20}C_{1} = 20$.Thus, the number of favorable outcomes $= 20 	imes 20 = 400$.The probability is given by $\frac{	ext{favorable outcomes}}{	ext{total outcomes}} = \frac{400}{780}$.Simplifying the fraction, we get $\frac{40}{78} = \frac{20}{39}$.

Out of $40$ consecutive integers, two integers are drawn at random. The probability that their sum is odd is

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