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Given that x is a hypergeometric random variable, compute P(x) for each of the following cases.

a. N=5, n=3, r=3, x=1

b. N=9, n=5, r=3, x=3

c. N=4, n=2, r=2, x=2

d. N=4, n=2, r=2, x=0

Short Answer

Expert verified

a. The Hypergeometric probability for N=5, n=3, r=3, x=1 is 0.30.

b. The Hypergeometric probability for N=9, n=5, r=3, x=3 is 0.11905.

c.The Hypergeometric probability for N=4, n=2, r=2, x=2 is 0.1666

d. The Hypergeometric probability for N=4, n=2, r=2, x=0 is 0.1666.

Step by step solution

01

Given Information

The x is a hypergeometric random variable.

02

State the Hypergeometric probability distribution

The Hypergeometric probability distribution is,

Px=rxN-rn-xNnx=Maximum0,n-N-r,...,Minimumr,n

03

(a) Compute the Hypergeometric probability for N=5, n=3, r=3, x=1

The Hypergeometric probability for N=5, n=3, r=3, x=1 is computed as:

Px=315-33-153=3!1!3-1!2!2!2-2!5!3!5-3!=3110=0.3

Hence, the required Hypergeometric probability for N=5, n=3, r=3, x=1 is 0.3.

04

(b) Compute the Hypergeometric probability for N=9, n=5, r=3, x=3

The Hypergeometric probability for N=9, n=5, r=3, x=3 is computed as:

Px=339-35-395=3!3!3-3!6!2!6-2!9!5!9-5!=115126=0.11905

Hence, the required Hypergeometric probability for N=9, n=5, r=3, x=3 is 0.11905.

05

(c) Compute the Hypergeometric probability for N=4, n=2, r=2, x=2

The Hypergeometric probability for N=4, n=2, r=2, x=2 is computed as:

Px=224-22-242=2!2!2-2!2!0!2-0!4!2!4-2!=116=0.1666

Hence, the required Hypergeometric probability for N=4, n=2, r=2, x=2 is 0.1666.

06

(d) Compute the Hypergeometric probability for N=4, n=2, r=2, x=0

The Hypergeometric probability for N=4, n=2, r=2, x=0 is computed as:

Px=204-22-042=2!0!2-0!2!2!2-2!4!2!4-2!=116=0.1666

Hence, the required Hypergeometric probability for N=4, n=2, r=2, x=0 is 0.1666.

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