Answer to Question #235663 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

Total Components = 12

Good Components = 7

Bad Components = 5

(a) That means either zero bad components or 1 bad components

Probability of zero bad components means all 4 are chosen from 7 good components.

Ways to choose 4 out of 7 $= C^7_4$

Total Ways to choose 4 out of 12 $= 12C^{12}_4$

Probability of one bad component means 3 are chosen from 7 good components. And 1 is chosen from 5 bad components

Ways to choose 3 out of 7 $= C^7_3$

Ways to choose 1 out of 5 $= C^5_1$

Total Ways to choose 4 out of 12 $= C^{12}_4$

$P(D≤1) = P(D=0) + P(D=1) \\ = \frac{\binom{5}{0} \binom{7}{4}}{\binom{12}{4}} + \frac{\binom{5}{1} \binom{7}{3}}{\binom{12}{4}} \\ = \frac{\frac{5!}{0!(5-0)!} \times \frac{7!}{4!(7-4)!} }{\frac{12!}{4!(12-4)!}} + \frac{\frac{5!}{1!(5-1)!} \times \frac{7!}{3!(7-3)!} }{\frac{12!}{4!(12-4)!}} \\ = \frac{1 \times 35}{495} + \frac{5 \times 35}{495} \\ = \frac{14}{33} \\ = 0.4242$

(b)

probability that Mr. Mechanical Engineer will have to get components from the box again means that atleast one is chosen from bad components.

$= 1 - \frac{C^5_0 \times C^7_4 }{ C^{12}_4} \\ P(D≥1) = 1 -P(D<1) \\ = 1 -P(D=0) \\ = 1 -\frac{\binom{5}{0} \binom{7}{4} }{\binom{12}{4}} \\ = 1 - \frac{\frac{5!}{0!(5-0)!} \times \frac{7!}{4!(7-4)!} }{ \frac{12!}{4!(12-4)!}} \\ = 1 - \frac{1 \times 35}{495} \\ = 1 -0.0707 \\ = 0.9293$