In: Statistics and Probability
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?
The Given problem is going to solved using Binomial distribution,
Binomial Distribution
If 'X' is the random variable representing the number of successes, the probability of getting ‘r’ successes and ‘n-r’ failures, in 'n' trails, ‘p’ probability of success ‘q’=(1-p) is given by the probability function
Given.
10 of the 200 units are known to have more than three major defects.
Probability of a unit to have than 3 defects = 10/200 = 0.05
If more than three major defects are found in any unit, they will reject the unit as defective
Therefore,
Probability of a unit to be defective : p = 0.05
q = 1-p = 1-0.05 = 0.95
Random sample of 8 units are selected by QA department;
Number of randomly selected units by QA department for inspection : n=8
Let X: Number of defective units of the 8 inspected units.
The Tower will be rejected if more than 2 of the 8 inspected units are defective i.e if X>2
Probability that the tower will be rejected = Probability that more than 2 of the 8 inspected units are defective = P(X>2)
Applying Binomial distribution,
X: Number of defective units of the 8 inspected units follows a Binomail distribution.
Probability of having 'r' defective units in 8 inspected units with probabity of unit being defective 'p = 0.05 and q =1 -p = 1-0.05 = 0.95 is given probability function
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 2)
P(X 2) = P(X=0) + P(X=1) +P(X=2)
P(X 2) = P(X=0) + P(X=1) +P(X=2) = 0.6634 + 0.2793 + 0.0515 = 0.9942
P(X>2) = 1-(P(X 2) = 1 - 0.9942 = 0.0058
Probability that the tower will be rejected = Probability that more than 2 of the 8 inspected units are defective = P(X>2) = 0.0058
Probability that the tower will be rejected = 0.0058