Probability & Statistics Ch.2 Section 2 Question 6

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Suppose a standard deck of 52 playing cards is thoroughly shuffled and a single card is drawn. Suppose an ace has value 1, a jack has value 11, a queen has value 12, and a king has value 13.

a) Compute $P(X = x)$ for every real number x, when X is the value of the card drawn. b) Suppose that Y = 1, 2, 3, or 4 when a diamond, heart, club, or spade is drawn. Compute $P(Y = y)$ for every real number y. c) Compute $P(W = w)$ for every real number w when $W = X + Y$.

Solution

(a) Compute $P(X = x)$ for every real number x

Each value (1 through 13) appears exactly 4 times in a 52-card deck:

$P(X=x)=0$ for all $x\notin\{1,2,3,4,5,6,7,8,9,10,11,12,13\}$

(b) Compute $P(Y = y)$ for every real number y

Each suit contains exactly 13 cards:

$P(Y=y)=0$ for all $y\notin\{1,2,3,4\}$

(c) Compute $P(W = w)$ for every real number w when $W = X + Y$

Since a single card determines both X and Y, we need to consider the probability of each (value, suit) combination.

For W = 2 (Ace of Diamonds): $P(W=2) = \frac{1}{52} \approx 0.019$

For W = 3 (Ace of Hearts or 2 of Diamonds): $P(W=3) = \frac{2}{52} \approx 0.038$

Similarly, for each value of W from 2 to 17:

• W can range from 2 (Ace=1 + Diamond=1) to 17 (King=13 + Spade=4)
• Each specific (value, suit) pair has probability $\frac{1}{52}$

The number of ways to get each sum W:

• W = 2, 3, 4, 5: 1, 2, 3, 4 ways respectively
• W = 6, 7, 8, 9, 10, 11, 12, 13, 14: 4 ways each
• W = 15, 16, 17: 3, 2, 1 ways respectively

$P(W=w)=0$ for all $w\notin\{2,3,4,...,17\}$