Out of 30 consecutive numbers,2 are chosen at | vedclass

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

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$\frac{15}{29}$

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(C) The total number of ways to choose $2$ integers from $30$ consecutive integers is given by $^{30}C_2 = \frac{30 	imes 29}{2} = 435$.For the sum of two numbers to be odd,one number must be even and the other must be odd.In $30$ consecutive integers,there are $15$ even numbers and $15$ odd numbers.The number of ways to choose one even number and one odd number is $^{15}C_1 	imes ^{15}C_1 = 15 	imes 15 = 225$.The required probability is $\frac{	ext{Favourable outcomes}}{	ext{Total outcomes}} = \frac{225}{435}$.Dividing both numerator and denominator by $15$,we get $\frac{225 \div 15}{435 \div 15} = \frac{15}{29}$.

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

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