Answer to Question #238589 in Statistics and Probability for James

Question

A box of a dozen mechanical components contains 7 good components and 5 bad components. Mr. Mechanical Engineer is preparing machine design for his four projects. One component is required per machine. He randomly selects 4 components from the box.

a. What is the probability that the Mr. Mechanical Engineer got at most one bad component?

b. What is the probability that Mr. Mechanical Engineer will have to get components from the box again?

Answer 1

a.

$$P(D\leq1)=P(D=0)+P(D=1)$$

$$=\dfrac{\dbinom{5}{0}\dbinom{7}{4}}{\dbinom{12}{4}}+\dfrac{\dbinom{5}{1}\dbinom{7}{3}}{\dbinom{12}{4}}$$

$$=\dfrac{\dfrac{5!}{0!(5-0)!}\cdot\dfrac{7!}{4!(7-4)!} }{\dfrac{12!}{4!(12-4)!}}$$

$$+\dfrac{\dfrac{5!}{1!(5-1)!}\cdot\dfrac{7!}{3!(7-3)!} }{\dfrac{12!}{4!12-4)!}}=$$

$$=\dfrac{1\cdot35}{495}+\dfrac{5\cdot35}{495}=\dfrac{14}{33}$$

b.

$$P(D\geq1)=1-P(D=0)$$

$$=1-\dfrac{\dbinom{5}{0}\dbinom{7}{4}}{\dbinom{12}{4}}=1-\dfrac{\dfrac{5!}{0!(5-0)!}\cdot\dfrac{7!}{4!(7-4)!} }{\dfrac{12!}{4!(12-4)!}}$$

$$=1-\dfrac{1\cdot35}{495}=\dfrac{98}{99}$$