Question

A quality control inspector has drawn a sample of 18 light bulbs from a recent production lot. If the number of defective bulbs is 2 or more, the lot fails inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will fail inspection? Round your answer to four decimal places.

Answer 1

There is a 98.58% probability that the lot will fail inspection.

Step-by-step explanation:

For each light bulb in the production lot, there are two possible outcomes. Either they are defective, or they are not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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.

In this problem we have that:

There are 18 light bulbs, so
`n = 18`.

30% of the bulbs in the lot are defective. This means that
`p = 0.3`.

If the number of defective bulbs is 2 or more, the lot fails inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will fail inspection?

This is
`P(X \geq 2)`.

Either there are 2 or more defective bulbs, or there are less than two. The sum of the probabilities of these events is decimal 1. So:

`P(X < 2) + P(X \geq 2) = 1`

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`

So

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

`P(X = 0) = C_(18,0).(0.30)^(0).(0.70)^(18) = 0.0016`

`P(X = 1) = C_(18,1).(0.30)^(1).(0.70)^(1t) = 0.0126`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.0016 + 0.0126 = 0.0142`

Finally

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0142 = 0.9858`

There is a 98.58% probability that the lot will fail inspection.

Answer 2

Answer: 0.9858

Step-by-step explanation:

Binomial distribution formula

`P(X=x)=^nC_x\ p^x\ (1-p)^(n-x)`, where P(x) is the probability of getting success in x trials , n is total number of trials and p is the probability of getting success in each trial.

Given : The probability that bulbs in the lot is defective = 0.30

Sample size = 18

If the number of defective bulbs is 2 or more, the lot fails inspection.

Then , the probability that the lot will fail inspection is given by :-

`P(X\geq2)=1-(P(X<2))\\\\=1-(P(0)+P(1))\\\\=1-(^(18)C_0\ (0.3)^0 (1-0.3)^(18)+^(18)C_1\ (0.3)^1 (1-0.3)^(17))\\\\=1-((0.3)^0 (0.7)^(18)+18(0.3)^1 (0.7)^(17))\approx0.9858`

Hence, the probability that the lot will fail inspection =0.9858