Question

A corporation receives a shipment having 18 pieces out of which 4 are substandard. If 6 pieces are selected as a random sample, and the shipment is rejected if any of these items proves to be substandard. Calculate the following if possible. If any part of question (you feel) is not possible to calculate give reason:
a. Develop the probability distribution table for the number of substandard pieces.
b. Calculate the expected number of substandard pieces and its variance.
c. What is the probability that less than 3 substandard pieces in the random sample?
d. Calculate the probability of accepting a shipment

Answer 1

a. Let $X=$ the number of substandard pieces among 6 taken pieces.

$$P(X=x)=\dfrac{\dbinom{4}{x}\dbinom{18-4}{6-x}}{\dbinom{18}{6}}$$

$$\dbinom{18}{6}=\dfrac{18!}{6!(18-6)!}=18564$$

$P(X=0)=\dfrac{\dbinom{4}{0}\dbinom{14}{6}}{18564}=\dfrac{3003}{18564}=\dfrac{11}{68}$

$P(X=1)=\dfrac{\dbinom{4}{1}\dbinom{14}{5}}{18564}=\dfrac{8008}{18564}=\dfrac{22}{51}$

$P(X=2)=\dfrac{\dbinom{4}{2}\dbinom{14}{4}}{18564}=\dfrac{6006}{18564}=\dfrac{11}{34}$

$P(X=3)=\dfrac{\dbinom{4}{3}\dbinom{14}{3}}{18564}=\dfrac{1456}{18564}=\dfrac{4}{51}$

$P(X=4)=\dfrac{\dbinom{4}{4}\dbinom{14}{2}}{18564}=\dfrac{91}{18564}=\dfrac{1}{204}$

$$\dfrac{11}{68} + \dfrac{22}{51} + \dfrac{11}{34} + \dfrac{4}{51} + \dfrac{1}{204}=1$$

$$\begin{matrix} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & \dfrac{11}{68} & \dfrac{22}{51} & \dfrac{11}{34} & \dfrac{4}{51} & \dfrac{1}{204} \end{matrix}$$

b,

$$E(X)=0\cdot\dfrac{11}{68}+1\cdot\dfrac{22}{51}+2\cdot\dfrac{11}{34}+3\cdot\dfrac{4}{51}+$$

$$+4\cdot\dfrac{1}{204}=\dfrac{4}{3}$$

$$E(X^2)=0\cdot\dfrac{11}{68}+1^2\cdot\dfrac{22}{51}+2^2\cdot\dfrac{11}{34}+3^2\cdot\dfrac{4}{51}+$$

$$+4^2\cdot\dfrac{1}{204}=\dfrac{128}{51}$$

$$Var(X)=E(X^2)-(E(X))^2=$$

$$=\dfrac{128}{51}-(\dfrac{4}{3})^2=\dfrac{112}{153}$$

c,

$$P(X<3)=P(X=0)+P(X=1)+P(X=2)=$$

$$\dfrac{11}{68} + \dfrac{22}{51} + \dfrac{11}{34} = \dfrac{11}{12}$$

$P(X<3)=\dfrac{11}{12}$

d.

$$P(accept)=P(X=0)=\dfrac{11}{68}$$