# Multiplication of conditional probabilities

Suppose 60% of people have a credit card, 37% of people have a debit card and 30% of people have neither a credit card or a debit card. Call these sets $A,B$ and $C$ respectively.

I am assisting a friend of mine and her solution online says that the probability that a random person have both a credit card and a debit card given they already have a credit card or a debit card is given by

$$P(A\cap B| A)P(A\cap B|B)$$

but I cannot fathom why this is the case, in particular, why it is the multiplication of these two probabilities. These two conditional probabilities do not seem like they are independent at all. My guess would have been that the probability is given by

$$P(A\cap B| A\cup B)$$

but I have tried to justify both of these solutions enough times that I've lost a lot of confidence. I understand that it is not a good practice to base things off of black box solutions, but I am normally able to convince myself otherwise if they are wrong and this time I can't. Any help would be appreciated.

• Forget your tries. Can you determine the probability that a random person have both a credit card and a debit card? – Did Dec 17 '13 at 14:45
• Well, that part is easy. It is just rearranging the equation $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. – JessicaK Dec 17 '13 at 14:50
• Yes, "easy" if you know P(A), P(B) and P(A∩B). You have P(A) and P(B) hence you need P(A∩B). How to get it? – Did Dec 17 '13 at 15:10
• I assume you mean $P(A\cup B) = 1 - P(C)$ – JessicaK Dec 17 '13 at 15:13
• Your friend's solution is coming out of the blue with no justification (and is wrong). Yours is simply the translation of what is being said: "the probability" (P of) "that a random person have both a credit card and a debit card" (A∩B) "given" (conditionally on) "they already have a credit card or a debit card" (A∪B). Well done. – Did Dec 17 '13 at 15:26

$P(A\cap B \mid A)P(A\cap B \mid B)$ is just wrong. Let's plug some numbers in it from a situation we can analyze completely by other means: flip a coin twice and let $A$ be "heads the first time" and $B$ be "heads the second time". Then obviously the probability of "heads both times" given that we know "heads at least once" is $1/3$, simply by counting which of the four possibilities are in each set. However, $P(A\cap B \mid A)P(A\cap B \mid B)$ evaluates to $1/4$, which is not the same as $1/3$ and therefore incorrect.
$P(A\cap B \mid A\cup B)$ is "right" in the sense that it is what you want to calculate, but as Did remarks, it is not really progress because it is just restating the word problem in symbols.
The people who have either debit or credit is exactly everyone who doesn't have nothing, that is, 70%, so the anwer is $$\frac{27}{70}$$