
Keith M. answered 06/09/15
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CMU Grad tutoring Mathematics and Computer Science
Although in general a pack of 52 playing cards may not be a standard deck, let's assume for this problem that the "usual" cards are present i.e. Ace through King in each of four suits (two red, two black). While some people do not consider the Ace to be a face card, I will include it (although the final answer is the same either way).
A pair of events is independent iff the probability of both events occurring is equal to the product of the probabilities of each event occurring (whether or not the other event occurs).
In drawing one card from the deck, the probability of drawing a black card P(F) is simply the number of black cards divided by the total number of cards.
P(F) = 26/52 = 1/2
The probability of drawing a face card P(G) is the number of face cards divided by the total number of cards.
P(G) = 16/52 = 4/13
The probability of drawing a black face card is the number of black face cards divided by the total number of cards.
P(F ∩ G) = 8/52 = 2/13
Now, we can answer the question:
Since P(F ∩ G) = 2/13 and P(F) × P(G) = 1/2 * 4/13 = 2/13, the events are independent.
Notice that this works because the proportion of face cards which are black is the same as the proportion of cards which are black. If this was not the case, then knowing whether you drew a face card would give you more information about whether you had drawn a back card, and vice versa. An extreme example would be to imagine a deck where half of the cards are black, but all of the face cards are black. In this case, P(F) × P(G) = P(G)/2, but P(F ∩ G) = P(G). In this case, the events are only independent if P(G) = 0, i.e. there are no face cards in the deck.


Keith M.
You're absolutely correct in the case that you draw multiple cards without replacement. In this problem, though, only a single card is drawn. The two "events" F and G refer to properties of that card.
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06/09/15
Mark M.
06/09/15