Let E F G denote three events Prove the following result If

Let E, F, G denote three events. Prove the following result: If E is independent of F, and E is independent of G, and F intersect G = naught, then E is independent of (F U G).

Solution

E is independent of F

p(E and F) = p(E)p(F)

E is independent of G

p(E and G) = p(E)p(G)

p(F and G) = 0

P(F U G) = p(F) + p(G)- p(F and G)

=p(F) + p(G)

p(EUFUG) = p(E) + p(FUG) -p(E and FUG)

=>p(E and FUG) = p(E) + p(FUG) -p(EUFUG)

= p(E) + p(F)+p(G) - [p(E) + p(F)+p(G) -p(E and F) - p(F and G)- p(E and G) + p(E and F and G)]

=p(E and F) + p(F and G) + p(E and G) - p(E and F and G)

=p(E)p(F) + 0 +p(E)p(G) - 0

= p(E) (P(F) + p(G) +0)

= p(E) (P(F) + p(G) -p(F and G))

= p(E) p(FUG)

hence proved