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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.

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  • $\begingroup$ Forget your tries. Can you determine the probability that a random person have both a credit card and a debit card? $\endgroup$ – Did Dec 17 '13 at 14:45
  • $\begingroup$ Well, that part is easy. It is just rearranging the equation $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. $\endgroup$ – JessicaK Dec 17 '13 at 14:50
  • $\begingroup$ 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? $\endgroup$ – Did Dec 17 '13 at 15:10
  • $\begingroup$ I assume you mean $P(A\cup B) = 1 - P(C)$ $\endgroup$ – JessicaK Dec 17 '13 at 15:13
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    $\begingroup$ 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. $\endgroup$ – Did Dec 17 '13 at 15:26
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$P(A\cap B \mid A)P(A\cap B \mid B)$ 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.

What you need to find is the proportion of people who have both debit and credit, but that's easy: 60% have credit (and possibly debit) and 30% have neither credit nor debit. The remaining 10% must have debit only; and since 37% have debit at all, that leaves 27% who have "debit but not only debit", that is "both debit and credit".

The people who have either debit or credit is exactly everyone who doesn't have nothing, that is, 70%, so the anwer is $$ \frac{27}{70} $$

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Ok let's split this up. It says 60% have a credit card, 37% have debit and 30% have neither but the probability doesn't add to 100% but why? Because some of those people who have credit cards also have debit cards. so finding out the percentage of how many people of the 60 percent have both multiplied by the percentage of many people of the 37% have both will give the total probability of those that have both. For example there are 5 people 2 have apples and 3 have oranges 2 have neither. So it's 2/3 x 3/3 = 2/3. Why? because since 2 have neither than the most that can have both is 3. For this example it would be .60/.7 x .37/.7. In other words P(A intersect B given A) x P(B intersect A given B) hope this helps

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