Answer

P(A U B) = P(A) + P(Ac ∩ B)?

Here is a simple but understandable approach. The method below is called the “counting argument” in combinatorics and probability. Since we have “two events” A & B, lets define the number of elements as follows :

A has ‘a’ elements. We write A = a for simplicity. Similarly,

A’ = a’ ( Ac, the number of elements in the complement of A )
B = b
B’ = b’
A n B = c
Total elements = y

Now A U B = a + b – c
So, P ( A U B ) = (No. of element in A U B )/ ( Total elements )
= ( a + b – c ) / y …(1)

Similarly, we establish that :
P(A) = a/y …(2) and

Consider (Ac n B). It contains the elements of B together with those that are in A’ or Ac. So, this is just the elements in B minus the elements that belong to both A and B. ( Draw a diagram or do an example and you will see this ),That is,
(Ac n B) = B – ( A n B ) = b – c, Thus,
P(Ac n B) = (b – c)/y…(3)

“Combine” (1), (2) and (3) to get,

P ( A U B ) = ( a + b – c ) / y
= (a/y) + (b -c)/y
= P(A) + P(Ac n B)

QED