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

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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