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(C) The total number of ways to choose $2$ integers from $30$ consecutive integers is given by $^{30}C_2 = \frac{30 	imes 29}{2} = 435$. In $30$ cons

Vedclass · Accepted Answer

$\frac{15}{29}$

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(C) The total number of ways to choose $2$ integers from $30$ consecutive integers is given by $^{30}C_2 = \frac{30 	imes 29}{2} = 435$. In $30$ consecutive integers,there are $15$ even numbers and $15$ odd numbers. The sum of two numbers is odd if and only if one number is even and the other is odd. The number of ways to choose one even number and one odd number is $^{15}C_1 	imes ^{15}C_1 = 15 	imes 15 = 225$. Therefore,the required probability is $P = \frac{225}{435} = \frac{15}{29}$.

Out of $30$ consecutive numbers,$2$ are chosen at random. The probability that their sum is odd,is

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