只需證明:P[(AUB)C]=P(AUB) *P(C).
P[(AUB)C]=P[ACUBC]= P(AC)+P(BC)- P[(AC)(BC)] (加法公式)
= P(AC)+P(BC)- P[(ABC)]
=P(A)*P(C) +P(B)*P(C) -P(A)*P(B)*P(C) (由已知條件得)
=[P(A) +P(B) -P(A)*P(B)]*P(C) (提出公因子)
=P(AUB)*P(C) (加法公式)
即確有:P[(AUB)C]=P(AUB) *P(C).
即知:AUB與C獨立