Question

Shipment of 7 television sets contains 3 defective sets. A hotel makes a random purchase of 3 of the sets. If x is the number of defective sets purchased by the hotel, find the probability distribution of X. Express the results graphically as a probability histogram.

Answer 1

Number of ways to select 3 televisions from 7:

$=\frac{7!}{3!(7-3)!}= \frac{5 \times 6 \times 7}{2 \times 3}= 35$

Number of ways to select 3 non-defective televisions from 4:

$= \frac{4!}{3!(4-3)!} =4$

The probability to select 3 non-defective:

$P(X=0) = \frac{4}{35}=0.1143$

Number of ways to select 1 defective television from 3:

$= \frac{3!}{1!(3-1)!} =\frac{2 \times 3}{2}= 3$

Number of ways to select 2 non-defective televisions from 4:

$= \frac{4!}{2!(4-2)!} =\frac{3 \times 4}{2}= 6$

The probability to select 1 defective:

$P(X=1) = \frac{3 \times 6}{35}=0.5143$

Number of ways to select 2 defective television from 3:

$= \frac{3!}{2!(3-2)!} = 3$

Number of ways to select 1 non-defective televisions from 4:

$= \frac{4!}{1!(4-1)!} = 4$

The probability to select 2 defective:

$P(X=2) = \frac{3 \times 4}{35}=0.3428$

Number of ways to select 3 defective television from 3:

$= \frac{3!}{3!(3-3)!} = 1$

The probability to select 3 defective:

$P(X=3) = \frac{ 1}{35}=0.0857$

Probability distribution