# Determining conditional probability when there are two events but one condition and vice versa

“Determine probability that a purchase is paid for by debit card OR credit card given that the customer paid over $100 for it” [2 events and 1 condition] “Compute the probability that a child survives an accident given that the child is either 3 OR 4 years old.” [ 1 condition, 2 events] My question is, how is this handled? What formula should I use? What is the notation? The OR really throws me off because I know OR means addition rule and in conditional probability, one must use multiplication rule. ## 1 Answer Paid by debit card$E_1$or credit card$E_2$is no different from one event$E$. Its probability is$P(E) = P(E_1) + P(E_2)$, since these are mutually exclusive. Then you can apply Bayes rule to$E$. Now, with conditional probability handy comes this formula:$P(F|E)P(E) = P(E|F)P(F) = P(E\cap F)$From it you can get (keeping in mind$E_1$and$ E_2$are mutually exclusive):$P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{P((E_1 \cup E_2)\cap F)}{P(F)} = \frac{P((E_1 \cap F)\cup(E_2 \cap F))}{P(F)} = \frac{P(E_1 \cap F) + P(E_2 \cap F)}{P(F)}=\frac{P(E_1 | F) P(F) + P(E_2 | F) P(F)}{P(F)} = P(E_1 | F) + P(E_2 | F)$which is quite obvious anyway because$E_1|F$and$E_2|F$are two mutually exclusive events, each a subset from the whole$E_1$and$E_2$, so should be additive. Now, doing it other way becomes trivial:$P(F|E) = \frac{P(E|F)P(F)}{P(E)} = \frac{(P(E_1 | F) + P(E_2 | F))P(F)}{P(E_1) + P(E_2)}$• Thanks for responding. I tried out what you said but I am stuck. I got to the point where P(E|F) [where F= customer paid over$100] becomes, by Bayes Rule: P(F|E)= P (E|F) X (F) / P(E). However, E, equals both events E1 and E2. P(E|F)= P(E and F)/ P(E). HOW do I get the probability for these 2 events and make them into 1 E? Do I take the probability of E1 and F and then the probability of E2 and F to get the total for the probability of E and F? I would appreciate some clarification. Thank you very much. Jul 13 '13 at 21:31
• @aboabo, Updated my answer. Jul 14 '13 at 1:31
• Thanks again for responding and explaining. I will continue this question later and hopefully would have understood it better. Jul 14 '13 at 23:11