Question

In a manufacturing process of resistors, it is assumed that 8.7% of the resistors produced are defective. If a random sample of 34 resistors is taken, what is the probability that there are exactly 3 defective resistors in the sample? Round your answer to the nearest 0.0001.

Answer 1

0.2345 = 23.45% probability that there are exactly 3 defective resistors in the sample

Explanation:

For each resistor, there are only two possible outcomes. Either they are defective, or they are not. The probability of a resistor being defective is independent of other resistors. So 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.

8.7% of the resistors produced are defective.

This means that
`p = 0.0870`

If a random sample of 34 resistors is taken, what is the probability that there are exactly 3 defective resistors in the sample?

This is P(X = 3) when n = 34. So

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

`P(X = 3) = C_(34,3).(0.087)^(3).(0.913)^(31) = 0.2345`

0.2345 = 23.45% probability that there are exactly 3 defective resistors in the sample