I have a problem with such simple task:
Probabilities of two independent events $A_1$ and $A_2$ are equal repectively $p_1$ and $p_2$. Find the probability that only one of the events will occur.
The answer given in the book is $P=p_1+p_2-2 \cdot p_1 \cdot p_2$
However I get the different answer. The result from above is got when such way of solving is applied:
$$P((A_1 \cap A_2') \cup (A_1' \cap A_2))=P(A_1 \cap A_2') + P(A_1' \cap A_2)$$
and so on. I don't udnerstand why there is no substraction of common part of events, as there is not mentioned that those events are mutually exclusive. Shouldn't the first step be:
$$P((A_1 \cap A_2') \cup (A_1' \cap A_2))=P(A_1 \cap A_2') + P(A_1' \cap A_2)- P((A_1 \cap A_2') \cap (A_1' \cap A_2))$$
?
Best Answer
The events $A_1\cap A_2'$ and $A_1'\cap A_2$ are mutually exclusive:
$$(A_1\cap A_2')\cap(A_1'\cap A_2)=(A_1\cap A_1')\cap(A_2\cap A_2')=\varnothing\cap\varnothing=\varnothing\;$$
You can certainly subtract $\Bbb P\big((A_1\cap A_2')\cap(A_1'\cap A_2)\big)$, but that’s $\Bbb P(\varnothing)=0$, so there’s no need to do so.
Since the events are independent, you then have $$\Bbb P(A_1\cap A_2')=p_1(1-p_2)$$ and $$\Bbb P(A_1'\cap A_2)=(1-p_1)p_2\;,$$ from which the result is a matter of a little algebra.