A shipment of 30 inexpensive digital watches, including 6 that are defective, is sent to a department store. The receiving department selects 10 at random for testing and rejects the whole shipment if 1 or more in the sample are found defective. What is the probability that the shipment will be rejected?

Let's notice that finding 1 or more defective watches in the sample is the complement (exactly the opposite) of finding 0 defective watches. That means P (1 or more defective watches) + P (0 defective watches) = 1

Let's find out P (0 defective watches):

The chances of getting 1 defective watch is 6 in 30 (in a shipment of 30 there are only 6 defective).

Therefore, we have 24/30 chances of getting a non-defective watch.

Needing 0 defective watches is the same as having 10 non defective watches.

When the receiving department select the first one, the probability of getting a non-defective watch is 24/30.

In the second one, they have a probability of 23/29 (they already took a non-defective one).

In the third one, the probability is 22/28.

And so on, up to the last one where the probability is 15/21.

As we need this to happen all at the same time, we have to multiply it:

P (0 defective watches) = [tex]\frac{24}{30}*\frac{23}{29}*\frac{22}{28}*\frac{21}{27}*\frac{20}{26}*\frac{19}{25}*\frac{18}{24}*\frac{17}{23}*\frac{16}{22}*\frac{15}{21}[/tex] = 0.065

Therefore, P (1 or more defective watches) = 1 - P (0 defective watches) = 1 - 0.065 = 0.935