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.

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:

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

Sample of 10

This means that

(a) exactly one defective item.

This is P(X = 1). So

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

(b) at most one defective item.

This is:

. So

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

step-by-step explanation:

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can you put the picture up?

step-by-step explanation: