n distinct items 1, 2, 3, dots, n are arrange | vedclass

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(D) Let $E_i$ be the event that the $i^{th}$ item is in the $i^{th}$ position.The probability that a specific set of $3$ items are in their correct po

Vedclass · Accepted Answer

$	ext{None of these}$

Vedclass · Answer

(D) Let $E_i$ be the event that the $i^{th}$ item is in the $i^{th}$ position.The probability that a specific set of $3$ items are in their correct positions is given by $\frac{(n-3)!}{n!}$.There are $\binom{n}{3}$ ways to choose $3$ positions out of $n$.For large $n$,the probability that exactly $k$ items are in their correct positions follows a Poisson distribution with parameter $\lambda = 1$,given by $P(X=k) = \frac{e^{-1}}{k!}$.The probability that at least $3$ items are in their correct positions is $1 - [P(X=0) + P(X=1) + P(X=2)]$.$P(X=0) = \frac{e^{-1}}{0!} = e^{-1}$,$P(X=1) = \frac{e^{-1}}{1!} = e^{-1}$,$P(X=2) = \frac{e^{-1}}{2!} = \frac{e^{-1}}{2}$.Sum $= e^{-1}(1 + 1 + 0.5) = 2.5 e^{-1} = \frac{2.5}{e} \approx \frac{2.5}{2.718} \approx 0.919$.Probability $= 1 - 0.919 = 0.081$,which is not equal to $\frac{1}{6} \approx 0.166$.Thus,the correct answer is $	ext{None of these}$.

$n$ distinct items $1, 2, 3, \dots, n$ are arranged in $n$ distinct positions $1, 2, 3, \dots, n$. What is the probability that at least three items are in their correct positions?

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