# If three cards are dealt from a deck of 52 playing cards, what is the probability that at least one will be an ace?

Q1 – If three cards are dealt from a deck of 52 playing cards, what is the probability that at least one will be an ace?

Q2- A store has identically appearing cases of canned vegetable, each containing 24 identically appearing cans, case A has 12 cans each of peas and beans, while B has 16 cans of peas and 8 cans of beans, A case is selected at random, and a cane is selected at random from that case, the cane when opened, is found to contain peas, what is the probability that case B was selected?

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For Q1, calculate the probability of the complement, i.e. that no card is an ace. For Q2, use conditional probability, i.e. $P(X | Y) = \frac {P(X \cap Y)}{P(Y)}$. –  Calvin Lin Dec 29 '12 at 10:05
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. –  Julian Kuelshammer Dec 29 '12 at 10:21
I want answer of these 2 questions in detail. –  Shabnam Dec 29 '12 at 10:35
Please ask one question per post, or explain the relationship between the questions. In the present case, I don't think there's a significant relationship between the questions; please remove one of them and ask it separately. –  joriki Dec 29 '12 at 11:08
@Shabnam, first of all you should give a try then ask for help. –  Ram Dec 29 '12 at 16:21

We will assume that there are precisely two cases. This is not absolutely clear from the wording. Essentially the same solution will work if we assume there are equal numbers of cases of each type. If possibly the numbers of cases of the two types are different, then the question cannot be answered.

Let $B$ be the event that Case B was selected, and let $P$ be the event the chosen can had peas. We want the probability of $B$, given that $P$ occurred. In symbols, we want $\Pr(B|P)$. By the standard formula for such things, we have $$\Pr(B|P)=\frac{\Pr(B\cap P)}{\Pr(P)}.\tag{1}$$ It remains to calculate the two probabilities on the right. We calculate the "harder" one, $\Pr(P)$.

The event $P$ can happen in two disjoint ways: (i) We picked Case A, and got a can of peas or (ii) We picked Case B, and got a can of peas.

We first calculate the probability of (i). With probability $\frac{1}{2}$, we picked Case A. Given that we picked Case A, the probability we got a can of peas is $\frac{12}{24}$. So the probability of (i) is $\frac{1}{2}\cdot \frac{12}{24}$.

Next we find the probability of (ii). By the same argument as the one for (i), the probability of (ii) is $\frac{1}{2}\cdot \frac{16}{24}$. It follows that $$\Pr(P)= \frac{1}{2}\cdot \frac{12}{24}+\frac{1}{2}\cdot \frac{16}{24}.$$

Next we find $\Pr(B\cap P)$. Conveniently, this has already been calculated, it is just the probability of (ii).

Finally, use Formula $(1)$.

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