Question

Assume that a deck of cards contains 4 aces, 5 other face cards, and 13 non-face cards, and that you randomly draw 3 cards. A random variable Z is defined to be 3 times the number of aces plus 2 times the number of other face cards drawn.

Complete the probability density function.

Answer 1

To find the probability density function of Z, we'll consider all possible values that Z can take and compute the probability of each.

Z can take values: 0 (no aces, no face cards), 2 (1 face card, no aces), 3 (1 ace, no other face cards), 4 (2 face cards, no aces), 6 (3 face cards or 2 aces), and 9 (3 aces). We now compute the probability for each case:

1. P(Z = 0): No aces or face cards.
We must draw all 3 cards from the 13 non-face cards: C(13,3)/C(22, 3) = 286/1540

2. P(Z = 2): One face card, no aces.
Choose 1 face card from the 5 available and 2 from the 13 non-face cards: [C(5, 1) * C(13, 2)]/C(22, 3) = 325/1540

3. P(Z = 3): One ace, no face cards.
Choose 1 ace from the 4 available and 2 from the 13 non-face cards: [C(4, 1) * C(13, 2)]/C(22, 3) = 260/1540

4. P(Z = 4): Two face cards, no aces.
Choose 2 face cards from the 5 available and 1 from the 13 non-face cards: [C(5, 2) * C(13, 1)] / C(22, 3) = 260/1540

5. P(Z = 6): Three face cards or 2 aces.
Three face cards: C(5, 3)/C(22, 3) = 10/1540
Two aces and one non-face card: [C(4, 2) * C(13, 1)]/C(22, 3) = 156/1540
Combined: (10 + 156)/1540 = 166/1540

6. P(Z = 9): Three aces.
Choose all 3 aces from the 4 available: C(4, 3)/C(22, 3) = 4/1540

So the probability density function of Z is:

P(Z) = {
9/1540, if Z = 0,
13/1540, if Z = 2,
26/1540, if Z = 3,
26/1540, if Z = 4,
83/1540, if Z = 6,
2/1540, if Z = 9,
0, otherwise
}