For $n$ independent events ${A_{i}^{\ }}^{'}s,\ P\left( A_{i} \right) = 1/(1 + i),$ $i = 1,\ 2,...,n$. The probability that atleast one of the events occurs is

Question

Dr. Abubacker · Accepted Answer

(c)
The probability $= P({\overline{A}}_{1} \cap {\overline{A}}_{2} \cap {\overline{A}}_{3} \cap \ldots. \cap {\overline{A}}_{n})$ (none of the events occur)
$= P\left( {\overline{A}}_{1} ight)P({\overline{A}}_{2})P\left( {\overline{A}}_{3} ight)\ldots P({\overline{A}}_{n})$
$= \frac{1}{2} \bullet \frac{2}{3} \bullet \frac{3}{4}\ldots\frac{n}{n + 1}$
$= \frac{1}{(n + 1)}\$
$	herefore$ Probability that atleast one of the events occurs
$= 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$

Q. For $n$ independent events ${A_{i}^{\ }}^{'}s,\ P\left( A_{i} \right) = 1/(1 + i),$ $i = 1,\ 2,...,n$. The probability that atleast one of the events occurs is

Explanation

Reviewed by Dr. Abubacker

Want more practice?