# Should it be A|B or B|A?

I came across a question in condition probability which left me wondering.

A departmental store reports that 30% of sales are transacted via cash, 60% are credit card and 10% with a debit card.

20% of the above cash purchases, 90% of the above credit card purchases and 80% of the above debit card purchases are made for purchases more than $200. So Let$C = {cash}$, D = {credit card}, E = {debit card} and M = {purchases >$200}

P(C) = 0.3
P(D) = 0.6
P(E) = 0.1


What I am not really sure is if it should be $P(M|C) = 0.2$ or is it $P(C|M) = 0.2$. Was wondering if I may get some advice on this please.

$P(M|C)$ denotes probability of the purchase being above $200 when you have observed that the purchase is through cash. You say 20% of the cash purchases are made for above$200$. So$P(M|C) = 0.2P(C|M)$is slightly more complicated. You observe that the purchase is above$200 and then you need the conditional probability that it is done by paying cash. You need to apply Bayes rule.

$$P(C|M) = \frac{P(C ) P(M|C)}{P(M)} = \frac{0.3 \times 0.2}{0.68} \approx 0.088$$

P.S: I must add, I am a newbie myself. So please get this verified by someone!

P.P.S: Sorry, as pointed out below, I made a small error with my normaliser. Fixed now.

• I think you may want to check the probablity of $M$, which IMAO should be $P(M)=P(M \cap C)+P(M \cap D)+P(M \cap E)$, from which follows that $P(M)=P(M|C)P(C)+P(M|D)P(D)+P(M|E)P(E)=.68$ so that $P(C|M)=.0882$. Commented Jan 26, 2014 at 21:56