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:
$p = 1-(p_1)(p_2)(p_3)…(p_n)$
Best Answer
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 $P(A)=1-P(A')$.
The actual complement of your event space is “at least one of the events occurs.”
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
$$\begin{align} P(E) &= P(E_1' \cap E_2' \cap E_3' \cap \cdots \cap E_n') \\ &= P(E_1')\,P(E_2')\,P(E_3')\cdots P(E_n') \\ &= (1-p_1)(1-p_2)(1-p_3)\cdots(1-p_n) \\ \end{align}$$