Question

1. (4 points) there are 8 computers in some university lab, of which 4 computers are defective and 4 computers are good. suppose 3 computers are selected at random in the lab, and let x be the number of defective computers from the sample. determine the probability distribution of the random variable x.

Explanation:

Step1: Determine the hyper - geometric distribution parameters

The hyper - geometric distribution formula is \(P(X = k)=\frac{\binom{K}{k}\binom{N - K}{n - k}}{\binom{N}{n}}\), where \(N\) is the population size, \(K\) is the number of success states in the population, \(n\) is the number of draws, and \(k\) is the number of observed successes.
Here, \(N = 8\) (total computers), \(K=4\) (defective computers), \(n = 3\) (computers selected).

Step2: Calculate \(P(X = 0)\)

\(P(X = 0)=\frac{\binom{4}{0}\binom{8 - 4}{3-0}}{\binom{8}{3}}\)
\(\binom{4}{0}=1\), \(\binom{4}{3}=\frac{4!}{3!(4 - 3)!}=4\), \(\binom{8}{3}=\frac{8!}{3!(8 - 3)!}=\frac{8\times7\times6}{3\times2\times1}=56\)
\(P(X = 0)=\frac{1\times4}{56}=\frac{1}{14}\)

Step3: Calculate \(P(X = 1)\)

\(P(X = 1)=\frac{\binom{4}{1}\binom{8 - 4}{3 - 1}}{\binom{8}{3}}\)
\(\binom{4}{1}=4\), \(\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=6\)
\(P(X = 1)=\frac{4\times6}{56}=\frac{3}{7}\)

Step4: Calculate \(P(X = 2)\)

\(P(X = 2)=\frac{\binom{4}{2}\binom{8 - 4}{3 - 2}}{\binom{8}{3}}\)
\(\binom{4}{2}=6\), \(\binom{4}{1}=4\)
\(P(X = 2)=\frac{6\times4}{56}=\frac{3}{7}\)

Step5: Calculate \(P(X = 3)\)

\(P(X = 3)=\frac{\binom{4}{3}\binom{8 - 4}{3 - 3}}{\binom{8}{3}}\)
\(\binom{4}{3}=4\), \(\binom{4}{0}=1\)
\(P(X = 3)=\frac{4\times1}{56}=\frac{1}{14}\)

Answer:

The probability distribution of \(X\) is:
\(P(X = 0)=\frac{1}{14}\), \(P(X = 1)=\frac{3}{7}\), \(P(X = 2)=\frac{3}{7}\), \(P(X = 3)=\frac{1}{14}\)