Question

A company generally purchases large lots of a certain kind of electronic device. A method is used that rejects a lot if 4 or more defective units are found in a random sample of 100 units. ​(a) What is the probability of rejecting a lot that is ​3% ​defective? ​(b) What is the probability of accepting a lot that is ​4% ​defective?

Answer 1

Using the binomial distribution, it is found that there is a:

a) 0.3526 = 35.26% probability of rejecting a lot that is 3% defective.

b) 0.4295 = 42.95% probability of accepting a lot that is 4% defective.

For each device, there are only two possible outcomes, either it is defective, or it is not. The probability of a device being defective is independent of any other device, hence the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• n is the number of trials.
• p is the probability of a success on a single trial.

In this problem, the sample has 100 units, hence $n = 100$.

Item a:

3% of the pieces are defective, hence $p = 0.03$.

The probability is:

$P(X \geq 4) = 1 - P(X < 4)$

In which:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{100,0}.(0.03)^{0}.(0.97)^{100} = 0.0476$

$P(X = 1) = C_{100,1}.(0.03)^{1}.(0.97)^{99} = 0.1471$

$P(X = 2) = C_{100,2}.(0.03)^{2}.(0.97)^{98} = 0.2252$

$P(X = 3) = C_{100,3}.(0.03)^{3}.(0.97)^{97} = 0.2275$

Then:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0476 + 0.1471 + 0.2252 + 0.2275 = 0.6474$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.6474 = 0.3526$

0.3526 = 35.26% probability of rejecting a lot that is 3% defective.

Item b:

4% of the pieces are defective, hence $p = 0.04$.

Lot is accepted if less than 4 units are defective, hence:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = 0) = C_{100,0}.(0.04)^{0}.(0.96)^{100} = 0.0169$

$P(X = 1) = C_{100,1}.(0.04)^{1}.(0.96)^{99} = 0.0703$

$P(X = 2) = C_{100,2}.(0.04)^{2}.(0.96)^{98} = 0.1450$

$P(X = 3) = C_{100,3}.(0.04)^{3}.(0.96)^{97} = 0.1973$

Then:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0169 + 0.0703 + 0.1450 + 0.1973 = 0.4295$

0.4295 = 42.95% probability of accepting a lot that is 4% defective.

A similar problem is given at brainly.com/question/24863377