Expert: Josh - 2/27/2008

Question
Suppose that you take all of the black cards out of a standard deck of 52 cards and thoroughly shuffle the remaining 26 red cards. From this deck of 26 red cards you will select 3 cards, one at a time, WITHOUT replacement, and record weather each card is picked is a face card (a jack, queen, or king), or not a face card.
a.) What is the probability that none of the 3 cards picked   in this way is a face card?
b.) what is the probability that exactly one of the three cards picked is a face card?

Answer
Hi Joshua,

It might be useful to draw some pictures initially until you get the hang of this.

Basic principles:

1. When solving this type of problem, we implicitly assume that each selection is done independently. That is, what happens next is NOT conditional upon previous outcomes. When this assumption holds, we are allowed to multiply the probability of particular events that we are interested in, round-by-round.
- e.g., if we want the probability of getting some pattern, like a face card in the first round (or pick) and an even number red card in the second pick, we consider these events SEQUENTIALLY. Suppose the probability of getting a face card is P(Face) in the first round, and the probability of getting an even red card is P(Even,Red) in the second round, the joint probability (i.e., having both things happening simultaneously) is given by the PRODUCT of P(Face) and P(Even,Red).

2. To actually find the answer numerically, we have to work out the probabilities such as P(Face) and P(Even,Red), using (i) counting techniques; or (ii) permutation/combination formulae with experience.

a) To avoid face cards at all times for three consecutive picks, we need to multiply P_1(Non-Face), P_2(Non-Face) and P_3(Non-Face) which represent the probability of picking non-face cards in round 1,2 and 3 respectively.

We begin with 26 red cards. Six of these are face cards counting the "jack(heart), queen(heart), king(heart)" and "jack(diamond), queen(diamond), king(diamond)".

In the first round, we have a 20 out of 26 chance of picking a non-face card. So, P_1(Non-Face)=20/26.
In the second round, with one non-face card already removed (without replacement), we have a 19 out of 25 chance of picking a non-face card. So, P_2(Non_Face) =19/25.
In the third round, with two non-face cards removed, we have P_3(Non-Face)=18/24.

Ans: Probability is (20/26)*(19/25)*(18/24).

b) This can happen in 3 ways.
Either Face, Non_Face and Non_Face,
or Non_Face, Face and Non_Face,
or Non_Face, Non_Face and Face.

The situation is symmetrical, it suffices to consider one case here.
For "Face, Non_Face and Non_Face", the probability is given by P_1(Face)*P_2(Non_Face)*P_3(Non_Face). By inspection, P_1(Face)=6/26. If you trace through the arguments we presented in part (a), you should arrive at P_2(Non_Face)=20/25, P_3(Non_Face)=19/24.

Because it can happen in 3 ways, the final answer is 3 x (6/26) x (20/25) x (19/24) = 0.4385