If Ai(i=1, 2, 3, ...,n) are n independent events with PAi=11+i for each i, the probability that none of Ai occurs is

P(Ai)=1i+1 PA1∩A2∩A3∩....An =P(A1)·P(A2)·P(A3)....P(An) =1-12 1-13 1-14....1-1n+1 =12·23·34.....nn+1=1n+1

If Ai(i=1, 2, 3, ...,n) are n independent events with PAi=11+i for each i, the probability that none of Ai occurs is

P(Ai)=1i+1 PA1∩A2∩A3∩....An =P(A1)·P(A2)·P(A3)....P(An) =1-12 1-13 1-14....1-1n+1 =12·23·34.....nn+1=1n+1

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

$\frac{1}{n + 2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} P (Ai) & = & \frac{1}{i + 1} \\ P (A_{1} \cap A_{2} \cap A_{3} \cap . . . . A_{n}) \\ = & P (A_{1}) \cdot P (A_{2}) \cdot P (A_{3}) . . . . P (A_{n}) \\ = & (1 - \frac{1}{2}) (1 - \frac{1}{3}) (1 - \frac{1}{4}) . . . . (1 - \frac{1}{n + 1}) \\ = & \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} . . . . . \frac{n}{n + 1} = \frac{1}{n + 1} \end{array}$