Question

Proof the the equation in the black writint using the equation in blue writing

ectio P(AUB) P(A)+P(B)-P(Ane). AUB P(AUB) P(A) +P(B) AAR

Answer 1

If we express the event $A\cup B$ as the union of two exclusive events $A$ and $A^{^{c}}\cap B$ .As is clear from the figure.

$A \cup B$ = $A \cup (A^{c \ca \ca} \cap B)$

Since the events on right hand side are exclusive, by axiom. 3 (i.e. P($A \cup B$) = P(A) + P(B) if A and B are mutually exclusive events)

$P(A \cup B) = P(A) + P ( A^{c} \cap B)$ ----------(1)

We know that,

$P ( A^{c} \cap B) = P (B) - P( A \cap B)$ ------------(2)

Putting the value of $P ( A^{c} \cap B)$ from (2) in (1), we get

$\therefore P(A \cup B) = P(A) + P (B) - P (A \cap B)$