For independent events A_1, A_2, dots, A_n, P | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The probability of non-occurrence of an event $A_i$ is given by $P(A_i^c) = 1 - P(A_i)$.Given $P(A_i) = \frac{1}{i + 1}$, we have $P(A_i^c) = 1 -

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The probability of non-occurrence of an event $A_i$ is given by $P(A_i^c) = 1 - P(A_i)$.Given $P(A_i) = \frac{1}{i + 1}$, we have $P(A_i^c) = 1 - \frac{1}{i + 1} = \frac{i}{i + 1}$.Since the events $A_1, A_2, \dots, A_n$ are independent, their complements $A_1^c, A_2^c, \dots, A_n^c$ are also independent.The probability that none of the events occur is the product of the probabilities of their non-occurrence:$P(	ext{none occur}) = P(A_1^c) \cdot P(A_2^c) \cdot \dots \cdot P(A_n^c)$$= \left( \frac{1}{2} ight) \cdot \left( \frac{2}{3} ight) \cdot \left( \frac{3}{4} ight) \dots \left( \frac{n}{n + 1} ight)$By canceling the terms in the numerator and denominator, we get:$= \frac{1}{n + 1}$.

For independent events $A_1, A_2, \dots, A_n$, $P(A_i) = \frac{1}{i + 1}$ for $i = 1, 2, \dots, n$. Then the probability that none of the events will occur is:

Similar Questions

From the word $POSSESSIVE$, a letter is chosen at random. The probability of it being $S$ is:

For two events $A$ and $B$,if $P(A) = P(A|B) = \frac{1}{4}$ and $P(B|A) = \frac{1}{2}$,then

If two regular six-sided dice are thrown,the probability that the sum will be a prime number is

Study the given information carefully to answer the question that follows. An urn contains $4$ green,$5$ blue,$2$ red,and $3$ yellow marbles. If $2$ marbles are drawn at random,what is the probability that both are red or at least $1$ is red?

If two balanced dice are tossed once,the probability of the event,that the sum of the integers coming on the upper sides of the two dice is $9$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module