Out of the given 25 consecutive positive inte | vedclass

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(A) Let the $25$ consecutive integers be $n, n+1, n+2, \dots, n+24$. Since $n$ is odd, the sequence contains $13$ odd numbers and $12$ even numbers. T

Vedclass · Accepted Answer

$\frac{289}{575}$

Vedclass · Answer

(A) Let the $25$ consecutive integers be $n, n+1, n+2, \dots, n+24$. Since $n$ is odd, the sequence contains $13$ odd numbers and $12$ even numbers. Total ways to choose $3$ integers from $25$ is $\binom{25}{3} = \frac{25 	imes 24 	imes 23}{3 	imes 2 	imes 1} = 2300$. The sum of three integers is even if: $1)$ All three are even: $\binom{12}{3} = \frac{12 	imes 11 	imes 10}{6} = 220$. $2)$ One is even and two are odd: $\binom{12}{1} 	imes \binom{13}{2} = 12 	imes \frac{13 	imes 12}{2} = 12 	imes 78 = 936$. Total favorable outcomes = $220 + 936 = 1156$. Probability = $\frac{1156}{2300} = \frac{289}{575}$.

Out of the given $25$ consecutive positive integers, three integers are drawn. If the least integer among the given $25$ integers is an odd number, then the probability that the sum of the three integers drawn is an even number is

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