Answer:
(a) The value of P (X = 2) is 0.3571.
(b) The value of P (X ≤ 1) is 0.5952.
Step-by-step explanation:
A Hypergeometric distribution is used to describe the probability distribution of x successes in n random draws from a population of size N that contains exactly r items that are considered as success. In this distribution each draw results in either a success or a failure.
The probability mass function of Hypergeometric distribution is:
[tex]P(X=x)=\frac{{r\choose x}{N-r\choose n-x}}{{N\choose n}}[/tex]
Given:
N = 9
r = 3
n = 4
(a)
Compute the value of P (X = 2) as follows:
[tex]P(X=2)=\frac{{3\choose 2}{9-3\choose 4-2}}{{9\choose 4}}=\frac{3\times 15}{126}=0.3571[/tex]
Thus, the value of P (X = 2) is 0.3571.
(b)
Compute the value of P (X ≤ 1) as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
[tex]=\frac{{3\choose 0}{9-3\choose 4-0}}{{9\choose 4}}+\frac{{3\choose 1}{9-3\choose 4-1}}{{9\choose 4}}\\=0.1190+0.4762\\=0.5952[/tex]
Thus, the value of P (X ≤ 1) is 0.5952.