Find the probability that none of the events occur?

Let the events $E_1, E_2, E_3, ... E_n$ be independent. $P\{E_K\} = p_k$. Find, $p$, the probability that none of the events occur.

The answer for this in the book is: $p = (1-p_1)(1-p_2)(1-p_3)...(1-p_n)$

Why is the answer not: $p = 1-(p_1)(p_2)(p_3)...(p_n)$

• The product of the $p_i's$ is the probability that all of the events occur so your formula computes the probability that at least one of the events does not occur. – lulu Sep 18 '17 at 23:03
• if one of the $p_i=0$ your p is going to be 1 in your formula it's absurd because of one of the event, all independant, it gives a certitude on p – Isham Sep 18 '17 at 23:09

Because $$(p_1)(p_2)(p_3)\cdots(p_n)=P(E_1\cap E_2 \cap E_3 \cap \cdots \cap E_n)$$
and the complement of your event space “none of the events occur” is not “all of the events occur,” so you cannot use $P(A')=(p_1)(p_2)(p_3)\cdots(p_n)$ in the formula $P(A)=1-P(A')$.
Your event $E$, “none of the events occur,” is equivalent to “$E_1$ does not occur and $E_2$ does not occur and $E_3$ does not occur and . . . and $E_n$ does not occur.”
Using $P(E_k')=1-p_k$, we have
\begin{align} P(E) &= P(E_1' \cap E_2' \cap E_3' \cap \cdots \cap E_n') \\ &= P(E_1')\,P(E_2')\,P(E_3')\cdots P(E_n') \\ &= (1-p_1)(1-p_2)(1-p_3)\cdots(1-p_n) \\ \end{align}