Question

A shipment of 6 computers contains three that are slightly defective. if a retailer receives three of these computers at random

Let X be the random variable representing the three slightly defective computers purchased by the retailer. Construction the Discrete Probability Distribution of the Random Variable X

Answer 1

Answer to Question #84210 – Math – Statistics and Probability

Question

A shipment of 6 computers contains three that are slightly defective. if a retailer receives three of these computers at random.

Let X be the random variable representing the three slightly defective computers purchased by the retailer. Construction the Discrete Probability Distribution of the Random Variable X.

Solution

We know that $\binom{3}{0} = \binom{3}{3} = 1$, $\binom{3}{1} = \binom{3}{2} = 3$, $\binom{6}{0} = \binom{6}{6} = 1$, $\binom{6}{1} = \binom{6}{5} = 6$, $\binom{6}{2} = \binom{6}{4} = 15$, $\binom{6}{3} = 20$

out of total 6 computers, 3 are defective, retailer receives 3 computers.

X = number of defective computers which retailer receives

X can be 0, 1, 2 or 3;

P (X=0) = P (0 defective and 3 non-defective) = $\frac{\binom{3}{0} \times \binom{3}{3}}{\binom{6}{3}} = \frac{1}{20}$

P (X=1) = P (1 defective and 2 non-defective) = $\frac{\binom{3}{1} \times \binom{3}{2}}{\binom{6}{3}} = \frac{3 \times 3}{20} = \frac{9}{20}$

P (X=2) = P (2 defective and 1 non-defective) = $\frac{\binom{3}{2} \times \binom{3}{1}}{\binom{6}{3}} = \frac{3 \times 3}{20} = \frac{9}{20}$

$$\mathrm{P} (\mathrm{X} = 3) = \mathrm{P} (3 \text{ defective and } 0 \text{ non-defective}) = \frac{\binom{3}{3} \times \binom{3}{0}}{\binom{6}{3}} = \frac{1}{20}$$

Probability distribution of random variable X

X = 0 1 2 3

P(x) = $\frac{1}{20}$ $\frac{9}{20}$ $\frac{9}{20}$ $\frac{1}{20}$

Answer provided by https://www.AssignmentExpert.com