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“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.

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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:

$P(F|E) = \frac{P(E|F)P(F)}{P(E)} = \frac{(P(E_1 | F) + P(E_2 | F))P(F)}{P(E_1) + P(E_2)}$

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  • $\begingroup$ 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. $\endgroup$
    – aboabo
    Jul 13 '13 at 21:31
  • $\begingroup$ @aboabo, Updated my answer. $\endgroup$
    – sashkello
    Jul 14 '13 at 1:31
  • $\begingroup$ Thanks again for responding and explaining. I will continue this question later and hopefully would have understood it better. $\endgroup$
    – aboabo
    Jul 14 '13 at 23:11

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