Answer to Question #126152 in Statistics and Probability for Sa’Mone Winters

Question

We have two situations:
Assignment 10
All Together, Now
a) Pull 5 cards from a standard deck, replacing and shuffling each time, and count the number of face cards.
b) Deal five cards from a standard deck (without replacement). Count the number of face cards.
1. Create probability distributions for both of these situations. Include your tables. [4]
2. Calculate the expected number of face cards for each situation. [2]
3. Explain which probability distribution type you picked for each, and why. [2, Communication]

Answer 1

a)

$X -$ random value, the number of faced cards

$P(X)\; -$ probability of getting $X$ faced cards with replacing

3.

Here is a binomial distribution. Experiment: shuffle a deck of cards, pull one card. Experiment repeats 5 times. Success is to pull faced card. So probability to pull $k$ faced cards after 5 experiments is $C_5^k \cdot p^k \cdot q^{5-k}$ , where $p = 12/52,\; q = 1-p=40/52$.

1.

$\def\arraystretch{1.6} \begin{array}{c|l} X & P(X) \\ \hline 0 & \left(\frac{40}{52}\right)^5 \approx 0.2693 \\ \hdashline 1 & 5 \cdot\left(\frac{12}{52}\right)^1\cdot\left(\frac{40}{52}\right)^4 \approx 0.404 \\ \hdashline 2 & 10\cdot\left(\frac{12}{52}\right)^2\cdot\left(\frac{40}{52}\right)^3 \approx 0.2424 \\ \hdashline 3 & 10\cdot\left(\frac{12}{52}\right)^3\cdot\left(\frac{40}{52}\right)^2 \approx 0.0727 \\ \hdashline 4 & 5 \cdot\left(\frac{12}{52}\right)^4\cdot\left(\frac{40}{52}\right)^1 \approx 0.0109 \\ \hdashline 5 & \left(\frac{12}{52}\right)^5 \approx 0.0007 \\ \end{array}$

2.

$E = 0\cdot0.2693 + 1\cdot0.404 + 2\cdot0.2424 + 3\cdot0.0727+4\cdot0.0109+5\cdot0.0007 \approx 1.154$

b)

$X -$ random value, the number of faced cards

$P(X)\;-$ probability of getting X faced cards without replacing

3.

Here is a hypergeometric distribution. Selections are made from two groups without replacing members of the groups. Success group consists of 12 elements, second group consists of 40 elements. So probability to pull $k$ elements from success group after 5 draws is $C_{12}^k \cdot C_{40}^{5-k}/C_{52}^5$ .

1.

$\def\arraystretch{1.6} \begin{array}{c|l} X & P(X) \\ \hline 0 & C_{12}^0 \cdot C_{40}^{5} / C_{52}^5 \approx 0.2532 \\ \hdashline 1 & C_{12}^1 \cdot C_{40}^{4} / C_{52}^5 \approx 0.422 \\ \hdashline 2 & C_{12}^2 \cdot C_{40}^{3} / C_{52}^5 \approx 0.2509 \\ \hdashline 3 & C_{12}^3 \cdot C_{40}^{2} / C_{52}^5 \approx 0.066 \\ \hdashline 4 & C_{12}^4 \cdot C_{40}^{1} / C_{52}^5 \approx 0.0076 \\ \hdashline 5 & C_{12}^5 \cdot C_{40}^{0} / C_{52}^5 \approx 0.0003 \\ \end{array}$

2.

$E = 0\cdot0.2532 + 1\cdot0.422 + 2\cdot0.2509 + 3\cdot0.066+4\cdot0.0076+5\cdot0.0003 \approx 1.1537$