Question

2.1-10. suppose there are 3 defective items in a lot (collection) of 50 items. a sample of size 10 is taken at random and without replacement. let x denote the number of defective items in the sample. find the probability that the sample contains (a) exactly one defective item. (b) at most one defective item.

Answer 1

a) 0.398 = 39.8% probability that the sample contains exactly one defective item.

b) 0.9021 = 90.21% probability that the sample contains at most one defective item.

Explanation:

The itens are chosen without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Suppose there are 3 defective items in a lot (collection) of 50 items

This means that
`k = 3, N = 50`

Sample of 10

This means that
`n = 10`

(a) exactly one defective item.

This is P(X = 1). So

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 1) = h(1,50,10,3) = (C_(3,1)*C_(47,9))/(C_(50,10)) = 0.398`

0.398 = 39.8% probability that the sample contains exactly one defective item.

(b) at most one defective item.

This is:

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

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 0) = h(0,50,10,3) = (C_(3,0)*C_(47,10))/(C_(50,10)) = 0.5041`

`P(X = 1) = h(1,50,10,3) = (C_(3,1)*C_(47,9))/(C_(50,10)) = 0.398`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5041 + 0.398 = 0.9021`

0.9021 = 90.21% probability that the sample contains at most one defective item.