If A1,A2,A3......An are independent events such that PAK=1K+1,K=1,2,3........n then the probability that none of the A1,A2,A3.......An occurs, is

pAK=1k+1→pA¯K=kk+1pA1C∩A2C∩A3C∩.....∩AnC=pA1C.pA2CpA3C....pAnC=12.23.34......nn+1=1n+1

If A1,A2,A3......An are independent events such that PAK=1K+1,K=1,2,3........n then the probability that none of the A1,A2,A3.......An occurs, is

pAK=1k+1→pA¯K=kk+1pA1C∩A2C∩A3C∩.....∩AnC=pA1C.pA2CpA3C....pAnC=12.23.34......nn+1=1n+1

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If $A_{1}, A_{2}, A_{3} ...... A_{n}$ are independent events such that $P (A_{K}) = \frac{1}{K + 1}, K = 1, 2, 3........ n$ then the probability that none of the $A_{1}, A_{2}, A_{3} ....... A_{n}$ occurs, is

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$\frac{1}{n + 1}$

$\frac{1}{n!}$

$\frac{n}{n + 1}$

$\frac{1}{{(n + 1)}^{2}}$

answer is C.

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Detailed Solution

$p (A K) = \frac{1}{k + 1}$

$\to p (\bar{A} K) = \frac{k}{k + 1}$

$p (A_{1}^{C} \cap A_{2}^{C} \cap A_{3}^{C} \cap ..... \cap A_{n}^{C})$

$= p (A_{1}^{C}) . p (A_{2}^{C}) p (A_{3}^{C}) .... p (A_{n}^{C})$

$= \frac{1}{2} . \frac{2}{3} . \frac{3}{4} ...... \frac{n}{n + 1} = \frac{1}{n + 1}$

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If A1,A2,A3......An are independent events such that PAK=1K+1,K=1,2,3........n then the probability that none of the A1,A2,A3.......An occurs, is

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If $A_{1}, A_{2}, A_{3} ...... A_{n}$ are independent events such that $P (A_{K}) = \frac{1}{K + 1}, K = 1, 2, 3........ n$ then the probability that none of the $A_{1}, A_{2}, A_{3} ....... A_{n}$ occurs, is