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

Answer:

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.

Step-by-step 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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 1) = h(1,50,10,3) = \frac{C_{3,1}*C_{47,9}}{C_{50,10}} = 0.398$

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

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,50,10,3) = \frac{C_{3,0}*C_{47,10}}{C_{50,10}} = 0.5041$

$P(X = 1) = h(1,50,10,3) = \frac{C_{3,1}*C_{47,9}}{C_{50,10}} = 0.398$

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

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