Christopher B. answered • 05/07/16
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Let E be the event that the second card is a diamond.
Let F1 be the event that the first card is the 5 of diamonds.
The formula for conditional probability is: P(E|F) = P(E∩F) / P(F).
It is easy to see that the probability of drawing one particular card out of 52 cards in a deck is (1/52). So, P(F) = (1/52).
The events E and F are not independent of each other. If the event F occurs then the probability of the event E occurring is altered.
E is the event defined to be drawing a diamond on the second card. This can occur if F has occurred beforehand, with probability (12/51). Let this event be E|F1. This can also occur if F has not occurred. There are two cases to be considered when F has not occurred. The first is that F has not occurred and that a diamond has not been drawn on the first card, let this event be E|F2. The second is that F has not occurred, yet the card drawn was some other card belonging to the suit of diamonds. Let this event be E|F3. I'll leave it to you as an exercise to make the associations between the conditions and the numbers.
Now, we can use the law of total probability to find P(E) = ((1/52)(12/51)) + ((39/52)(13/51)) + ((13/52)(12/51)) = (675/2652).
However, we are not concerned with all of the cases in which the second card drawn will have diamonds as its suit. So, simply by nullifying the other cases we obtain,
(P(E|F1) P(F1))/P(F1)= (1/52)(12/51)*52 = 12/51 = 4/17.
Yes, it took me a while to get to the crux of the problem and it is often possible to solve these problems intuitively, but I thought it may be educational to demonstrate some concepts from probability theory along the way. Practicing formality on the easier problems will make it easier to execute it on the more difficult problems.
