Question

A company that makes rock and roll t-shirts has a printing problem that causes the band names to smear on 4% of the t-shirts. The daily production run is 2,000 t-shirts. What is the probability, if a sample of 100 t-shirts is checked, there will be smeared names on at most 4 t-shirts?

Answer 1

62.89% probability that there will be smeared names on at most 4 t-shirts

Explanation:

For each shirt, there are only two possible outcomes. Either there will be smeared names, or there will not. The probability of there being smeared names on a shirt is independent of other shirts. 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.

4% of the shirts are smeared:

So
`p = 0.04`

What is the probability, if a sample of 100 t-shirts is checked, there will be smeared names on at most 4 t-shirts?

This is
`P(X \leq 4)` when
`n = 100`. So

`P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)`

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

`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`

`P(X = 4) = C_(100,4).(0.04)^(4).(0.96)^(96) = 0.1994`

`P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0169 + 0.0703 + 0.1450 + 0.1973 + 0.1994 = 0.6289`

62.89% probability that there will be smeared names on at most 4 t-shirts