If A1- A2- - An are n independent events such that PAi-1i-1- i-1- 2- - n- The probability that none of A1- A2- - An occurs is

The correct options are A less than 1n C 1n+1 P(A'∩ A'_2 ....∩ A'_n) =P(A'_1)P(A'_2)...P(A'_n) [∵ A_1, A_2, ..., A_n are independent] ...

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Standard XII

Mathematics

Mutually Exclusive Events

If A1, A2, ...

Question

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

\frac{n}{n + 1}

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\frac{1}{n + 1}

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less than

\frac{1}{n}

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n o n e o f t h e s e

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Solution

The correct options are
A less than $\frac{1}{n}$
C $\frac{1}{n + 1}$
$P (A^{'} \cap A_{2}^{'} . . . . \cap A_{n}^{'})$
$= P (A_{1}^{'}) P (A_{2}^{'}) . . . P (A_{n}^{'})$
$[∵ A_{1}, A_{2}, . . ., A_{n}$ are independent]
$= (1 - \frac{1}{2}) (1 - \frac{1}{3}) . . . . (1 - \frac{1}{n + 1})$
$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times . . . \times \frac{n - 1}{n} \times \frac{n}{n + 1} = \frac{1}{n + 1} < \frac{1}{n}$ .

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Similar questions

Q. Assertion: Given

E_{i} (i = 1, 2, . . . . . . . n)

are

n

independent events such that

P (_{i}) = \frac{i}{i + 1}, 1 \leq i \leq n

, then the probability that none of the

n

events occur is

\frac{n}{n + 1}

Reason: Probability of occurence of all independent events is equal to the product of the probabilities of individual events.

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If A1,A2,....,An are n independent events such that P(Ai)=1i+1,i=1,2,...,n. The probability that none of A1,A2,....An occurs is

The correct options are A less than 1n C 1n+1P(A′∩A′2....∩A′n)=P(A′1)P(A′2)...P(A′n)[∵A1,A2,...,An are independent]=(1−12)(1−13)....(1−1n+1)=12×23×34×...×n−1n×nn+1=1n+1<1n.

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