Answer to Question #149675 in Statistics and Probability for Randal Rodriguez

Question

 A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets. Considering that y is the number of non-defective sets purchased by the hotel

(a) What are the values for the set of y?

(b) What is the probability of purchasing 2 non-defective television sets? (Fraction form)

(c) make a probability distribution table.

Answer 1

(a) Since a shipment of 7 television sets contains 2 defective sets and a hotel makes a random purchase of 3 of the sets,  for the number $y$ of non-defective sets purchased by the hotel we have $y\in\{1,2,3\}$.

(b) The probability $p$ of purchasing 2 non-defective television sets is

$p=\frac{{5 \choose 2}{2\choose 1}}{7 \choose 3}=\frac{\frac{5!}{2!\cdot 3!}\cdot 2}{\frac{7!}{3!\cdot 4!}}=\frac{5!\cdot 4!\cdot 2}{2!\cdot 7!}=\frac{2\cdot 3\cdot 4}{6\cdot 7}=\frac{4}{7}.$

(c) Let us make a probability distribution table for $y\in\{1,2,3\}$.

We already have that $P(y=2)=\frac{4}{7}$.

$P(y=1)=\frac{{5 \choose 1}{2\choose 2}}{7 \choose 3}=\frac{5}{\frac{7!}{3!\cdot 4!}}=\frac{5\cdot 3!\cdot 4!}{ 7!}=\frac{5\cdot 2\cdot 3}{7\cdot 6\cdot 5}=\frac{1}{7}$

$P(y=3)=\frac{{5 \choose 3}}{7 \choose 3}=\frac{\frac{5!}{3!\cdot 2!}}{\frac{7!}{3!\cdot 4!}}=\frac{5!\cdot 4!}{2!\cdot 7!}=\frac{ 2\cdot 3\cdot 4}{2\cdot 7\cdot 6}=\frac{2}{7}$

Therefore, a probability distribution table is the following:

$\begin{array}{|c|c|c|c|} \hline y & 1 & 2 & 3\\ p & \frac{1}{7} & \frac{4}{7} & \frac{2}{7} \\ \hline \end{array}$