Probability & Statistics Ch.2 Section 3 Question 17

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

An urn contains 4 black balls and 5 white balls. After a thorough mixing, a ball is drawn from the urn, its color is noted, and the ball is set aside. The remaining balls are then mixed and a second ball is drawn.

a) What is the probability distribution of the number of black balls observed? b) What is the probability distribution of the number of white balls observed?

Solution

Since the trials are dependent (no replacement), we use the Hypergeometric distribution.

(a) What is the probability distribution of the number of black balls observed?

From Example 2.3.7 in the textbook:

p_X(x) = P(X=x) = \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}

Where:

N = 9 (total number of balls)
M = 4 (number of black balls)
N - M = 5 (number of white balls)
n = 2 (number of balls drawn)
x = number of black balls drawn

X \sim \text{Hypergeometric}(9, 4, 2)

P(X=x) = \frac{\binom{4}{x}\binom{5}{2-x}}{\binom{9}{2}}

(b) What is the probability distribution of the number of white balls observed?

This is the same as above with M and N-M swapped:

N = 9 (total number of balls)
M = 5 (number of white balls - now the “desired” type)
N - M = 4 (number of black balls)
n = 2 (number of balls drawn)
x = number of white balls drawn

X \sim \text{Hypergeometric}(9, 5, 2)

P(X=x) = \frac{\binom{5}{x}\binom{4}{2-x}}{\binom{9}{2}}