Question

The probability of n independent events are $p_1,p_2,\cdots p_n.$ Find an expression for the probability that at least one of the events will happen.

• A. $1-\left(p_1\right)\left(p_2\right)\cdots \left(p_n\right)$
• B. $(1-p_1)(1-p_2)\cdots (1-p_n)$
• C. $1-(1-p_1)(1-p_2)\cdots (1-p_n)$
• D. $\left(p_1\right)\left(p_2\right)\cdots \left(p_n\right)$

Answer 1

Answer: (C)

$1-(1-p_1)(1-p_2)\cdots (1-p_n)$

Let $E_1,E_2,....E_n$ be n independent events.
Given probabilities of these events as $p_1,p_2,....p_n$
Since, probability of happening of $E_1$ is $p_1$
Then probability of non-happening of event $E_1$ is $1-p_1$
Similarly, probability of non-happening of event $E_2$ is $1-p_2$
Similarly, probability of non-happening of event $E_n$ is $1-p_n$
Now, probability of non-happening of any of the events is $(1-p_1)(1-p_2)....(1-p_n)$
So, probability of happening of at least one event is $1-(1-p_1)(1-p_2)....(1-p_n)$