Answer to Question #151679 in Statistics and Probability for donabel

Question

A shipment of 15 similar microcomputers to a retail outlet contains 5
that are defective. If a school makes a random purchase of 4 of these
computers, find the probability distribution and the expectation for the
number of defective items.

Answer 1

Let $X=$ the number of defective computers purchased. 

There are $15-5=10$  working microcomputers.

There are $^{15}C_4=\dbinom{15}{4}=\dfrac{15!}{4!(15-4)!}=\dfrac{15(14)(13)(12)}{1(2)(3)(4)}=1365$ ways to select 4 computers.

$$P(X=0)=\dfrac{^{5}C_0\cdot^{10}C_4}{^{15}C_4}=\dfrac{1\cdot210}{1365}=\dfrac{2}{13}$$

$$P(X=1)=\dfrac{^{5}C_1\cdot^{10}C_3}{^{15}C_4}=\dfrac{5\cdot210}{1365}=\dfrac{40}{91}$$

$$P(X=2)=\dfrac{^{5}C_2\cdot^{10}C_2}{^{15}C_4}=\dfrac{10\cdot45}{1365}=\dfrac{30}{91}$$

$$P(X=3)=\dfrac{^{5}C_3\cdot^{10}C_1}{^{15}C_4}=\dfrac{10\cdot10}{1365}=\dfrac{20}{273}$$

$$P(X=4)=\dfrac{^{5}C_4\cdot^{10}C_0}{^{15}C_4}=\dfrac{5\cdot1}{1365}=\dfrac{1}{273}$$

Check

$$\dfrac{2}{13}+\dfrac{40}{91}+\dfrac{30}{91}+\dfrac{20}{273}+\dfrac{1}{273}=1$$

$$\begin{matrix} x & 0 & 1 & 2 & 3 & 4 \\ \\ p(x) & \dfrac{2}{13} & \dfrac{40}{91} & \dfrac{30}{91} & \dfrac{20}{273} & \dfrac{1}{273} \end{matrix}$$

$$E(X)=0\cdot\dfrac{2}{13}+1\cdot\dfrac{40}{91}+2\cdot\dfrac{30}{91}+3\cdot\dfrac{20}{273}$$

$$+4\cdot\dfrac{1}{273}=\dfrac{4}{3}$$

$$E(X)=\dfrac{4}{3}$$