A question about pairwise mutual exclusive events There is a common problem in the field of probability as follows:
"On a vessel containing 3 white and 5 black balls, 4 balls are transferred into an empty vessel. From this vessel a ball is drawn and it is found to be white. What is the probability that out of 4 balls transferred 3 are white and 1 is black?"
The solution is given on page 335 of the book "Engineering Mathematics by Sarveswara Rao Koneru (Universities Press, 2002)" as follows:
"The four balls transferred can be only one of the following combinations:
Event $A_1$ : 4B + 0W
Event $A_2$ : 3B + 1W
Event $A_3$ : 2B + 2W
Event $A_4$ : 1B + 3W
$A_1$, $A_2$, $A_3$ and $A_4$ are pairwise mutual exclusive since just one of them must happen so if E is the event of choosing a white ball from the second vessel,
$E = EA_1 \cup EA_2 \cup EA_3 \cup EA_4$
(and the answer continues to the end...)"
My question is how does the above equation hold and how does the author concluded that?
 A: $\{A_1,A_2,A_3,A_4\}$ are not just pairwise mutually exclusive events, they are also exhaustive.  This is a set of events which partitions the outcome set.
Let $\Omega$ be the set of all possible outcomes for drawing four balls from the vessel.   Since any outcome must be in one among the events from $\{A_1,A_2,A_3,A_4\}$, then $\Omega=A_1\cup A_2\cup A_3\cup A_4$.   Thus they are exhaustive over $\Omega$.
This exhaustive property is what is used to claim:
$$\begin{align}E&=E\cap\Omega\\&=E\cap(A_1\cup A_2\cup A_3\cup A_4)\\&=(E\cap A_1)\cup(E\cap A_2)\cup(E\cap A_3)\cup(E\cap A_4)\end{align}$$

Mutual exclusivity comes next.
Since any event in $E$, is an event in $\Omega$, it  must be in exactly one among the events from $\{A_1,A_2,A_3,A_4\}$, and therefore in exactly one among the events from $\{(E\cap A_1),(E\cap A_2),(E\cap A_3),(E\cap A_4)\}$.  Therefore this later set of events is also pairwise mutually exclusive.   The probability of a union of pairwise mutually exclusive events equals the sum of the probabilities of those events.
As we have already shown that this union equals $E$, thus:
$$\mathsf P(E)=\mathsf P(E\cap A_1)+\mathsf P(E\cap A_2)+\mathsf P(E\cap A_3)+\mathsf P(E\cap A_4)$$
