Question

10. A large shipment of video streaming devices is accepted upon delivery if an inspection of 20 randomly selected video streaming devices yields no more than 1 defective item. a. Find the probability that the shipment is accepted if 3% of the total shipment is defective. b. Find the probability that the shipment is accepted if 15% of the total shipment is defective.

Answer 1

a) 0.8802 = 88.02% probability that the shipment is accepted

b) 0.1756 = 17.56% probability that the shipment is accepted if 15% of the total shipment is defective.

Explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of any other item. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

20 randomly selected video streaming devices

This means that
`n = 20`

a. Find the probability that the shipment is accepted if 3% of the total shipment is defective.

3% defective means that
`p = 0.03`

Probability of at most 1 defective is:

`P(X \leq 1) = P(X = 0) + P(X = 1)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.03)^(0).(0.97)^(20) = 0.5438`

`P(X = 1) = C_(20,1).(0.03)^(1).(0.97)^(19) = 0.3364`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5438 + 0.3364 = 0.8802`

0.8802 = 88.02% probability that the shipment is accepted.

b. Find the probability that the shipment is accepted if 15% of the total shipment is defective

Now
`p = 0.15`. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.15)^(0).(0.85)^(20) = 0.0388`

`P(X = 1) = C_(20,1).(0.15)^(1).(0.85)^(19) = 0.1368`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756`

0.1756 = 17.56% probability that the shipment is accepted if 15% of the total shipment is defective.