Answer to Question #248904 in Mechanical Engineering for pari

Question

A high-rise tower built by a developer contains 200 condominium units. The QA department of the developer has to approve the tower before turning it over to sales. The QA department will select a random sample of 8 units and inspect them. If more than three major defects are found in any unit, they will reject the unit as defective. If more than 2 of the 8 inspected units are defective, the entire tower will be rejected. If 10 of the 200 units are known to have more than three major defects, what is the probability that the tower will be rejected?

Answer 1

Probability of a unit to have more than 3 defects=$\frac{10}{200}=0.05$

Probability of a unit to be defective, p=0.05

q=1-p=1-0.05=0.95

Number of randomly selected inspected units, n=8

Let x be number of defective units of the inspected 8

Tower will be rejected if X>2

Probability that tower us rejected P(X>2)

Apply Binomial distribution

$P(X=r)=\dbinom{8}{r}*0.05^{r}*0.95^{8-r}$

Probability that the tower will be rejected=probability that more than 2 of the 8 inspected units are defective=P(X>2)

P(X>2)=1-P(X$\le$ 2)

P(X$\le$ 2)=P(X=0)+P(X=1)+P(X=2)

$P(X=0)=\dbinom{8}{0}*0.05^{0}*0.95^{8-0}$

=1*1*0.6634=0.6634

$P(X=1)=\dbinom{8}{1}*0.05^{1}*0.95^{8-1}$

=8*0.05*0.6983

=0.2793

$P(X=2)=\dbinom{8}{2}*0.05^{2}*0.95^{8-2}$

=28*0.0025*0.7351

=0.0515

P(X$\le$ 2)=0.6634+0.2793+0.0515=0.9942

P(X>2)=1-P(X$\le$ 2)=1-0.9942

=0.0058

Probability that tower will be rejected=0.0058