Question

A certain manufacturing process yields electrical fuses of which, in the long run

15% are defective. Find the probability that in a random sample of size n=10, fuses
selected from this process, there will be
(i) No defective fuse
(ii) At least one defective fuse
(iii) Exactly two defective fuses
(iv) At most one defective fuse

Answer 1

i) 0.1969 = 19.69% probability that there will be no defective fuse.

ii) 0.8031 = 80.31% probability that there will be at least one defective fuse.

iii) 0.2759 = 27.59% probability that there will be exactly two defective fuses.

iv) 0.5443 = 54.43% probability that there will be at most one defective fuse.

Explanation:

For each fuse, there are only two possible outcomes. Either it is defective, or it is not. The probability of a fuse being defective is independent of any other fuse, which means that the binomial probability distribution is used 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.

15% are defective.

This means that
`p = 0.15`

We also have:

`n = 10`

(i) No defective fuse

This is
`P(X = 0)`. So

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

`P(X = 0) = C_(10,0).(0.15)^(0).(0.85)^(10) = 0.1969`

0.1969 = 19.69% probability that there will be no defective fuse.

(ii) At least one defective fuse

`P(X \geq 1) = 1 - P(X = 0)`

We already have P(X = 0) = 0.1969, so:

`P(X \geq 1) = 1 - 0.1969 = 0.8031`

0.8031 = 80.31% probability that there will be at least one defective fuse.

(iii) Exactly two defective fuses

This is P(X = 2). So

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

`P(X = 2) = C_(10,2).(0.15)^(2).(0.85)^(8) = 0.2759`

0.2759 = 27.59% probability that there will be exactly two defective fuses.

(iv) At most one defective fuse

This is:

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

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

`P(X = 0) = C_(10,0).(0.15)^(0).(0.85)^(10) = 0.1969`

`P(X = 1) = C_(10,1).(0.15)^(1).(0.85)^(9) = 0.3474`

Then

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1969 + 0.3474 = 0.5443`

0.5443 = 54.43% probability that there will be at most one defective fuse.