Answer to Question #342794 in Statistics and Probability for Ahmad

Question

Suppose that the racial/ethnic distribution in a large city is given by the table that follows.

Black Hispanic Other

20% 15% 65%

Suppose that a jury of twelve members is chosen from this city in such a way that each resident has an equal probability of being selected independently of every other resident.

Lets find probability that the jury contains:

a. three Black, two Hispanic, and seven other members.

b. four Black and eight other members.

c. at most one Black member.

Answer 1

To solve this problem, let X=(X1,X2,X3)

where X1= number of Black members, X2= number of Hispanic members, and X3= number of Other members. Then X has a multinomial distribution with parameters n=12 and π=(0.20, 0.15, 0.65).

a. The answer to the first part is

$P(X_1=3, X_2=2, X_3=7)=\frac{n!}{x_1!x_2!x_3!}\pi_1^{x_1}\pi_2^{x_2}\pi_3^{x_3}=\frac{12!}{3!2!7!}(0.20)^3(0.15)^2(0.65)^7=0.0699$

b. The answer to the second part is

$P(X_1=4, X_2=0, X_3=8)=\frac{n!}{x_1!x_2!x_3!}\pi_1^{x_1}\pi_2^{x_2}\pi_3^{x_3}=\frac{12!}{4!0!8!}(0.20)^4(0.15)^0(0.65)^8=0.0252$

c. For the last part, note that "at most one Black member" means X1=0 or X1=1. X1 is a binomial random variable with n=12 and π1=0.2. Using the binomial probability distribution,

$P(X_1=0)=\frac{12!}{0!12!}(0.20)^0(0.8)^{12}=0.0687$

and

$P(X_1=1)=\frac{12!}{1!11!}(0.20)^1(0.8)^{11}=0.2061$

Therefore, the answer is:

$P(X_1=0)+P(X_1=1)=0.0687+0.2061=0.2748$