Out of 11 consecutive natural numbers,if thre | vedclass

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(C) Let the $11$ consecutive natural numbers be $n, n+1, n+2, \dots, n+10$.The total number of ways to select $3$ numbers out of $11$ is given by ${}^

Vedclass · Accepted Answer

$\frac{5}{33}$

Vedclass · Answer

(C) Let the $11$ consecutive natural numbers be $n, n+1, n+2, \dots, n+10$.The total number of ways to select $3$ numbers out of $11$ is given by ${}^{11}C_{3} = \frac{11 	imes 10 	imes 9}{3 	imes 2 	imes 1} = 165$.For three numbers $a, b, c$ to be in $A.P.$,they must satisfy $a + c = 2b$. This implies that $a + c$ must be an even number.$a + c$ is even if both $a$ and $c$ are even or both $a$ and $c$ are odd.Case $1$: $11$ consecutive numbers contain $6$ even and $5$ odd numbers.Number of ways to choose $2$ even numbers is ${}^{6}C_{2} = 15$.Number of ways to choose $2$ odd numbers is ${}^{5}C_{2} = 10$.Total favourable cases $= 15 + 10 = 25$.Case $2$: $11$ consecutive numbers contain $5$ even and $6$ odd numbers.Number of ways to choose $2$ even numbers is ${}^{5}C_{2} = 10$.Number of ways to choose $2$ odd numbers is ${}^{6}C_{2} = 15$.Total favourable cases $= 10 + 15 = 25$.In both cases,the number of favourable outcomes is $25$.Therefore,the probability $P = \frac{25}{165} = \frac{5}{33}$.

Out of $11$ consecutive natural numbers,if three numbers are selected at random (without repetition),then the probability that they are in $A.P.$ with a positive common difference is:

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