A. Four factories are to be randomly selected from a list of eleven factories for a full health and

safety inspection
i. Calculate the number of ways that four factories can be selected.
ii. Two of the factories have the same owner. Calculate the probability that both of these
factories with the same owner will be included in the inspection.

A. Four factories are to be randomly selected from a list of eleven factories for a full health and

safety inspection
i. Calculate the number of ways that four factories can be selected.
ii. Two of the factories have the same owner. Calculate the probability that both of these
factories with the same owner will be included in the inspection.

i. To calculate the number of ways that four factories can be selected from a list of eleven factories, we can use the combination formula. The formula for combination is:

C(n, r) = n! / (r!(n - r)!)

Where n is the total number of items to choose from, and r is the number of items to choose.

In this case, we have 11 factories to choose from and we want to select 4 factories. Applying the combination formula, we get:

C(11, 4) = 11! / (4!(11 - 4)!) = 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330

So, there are 330 different ways to select four factories from a list of eleven factories.

ii. To calculate the probability that both factories with the same owner will be included in the inspection, we need to consider that one factory is already selected (since they have the same owner). We then need to select three more factories from the remaining ten (excluding the one already selected).

The probability of selecting a specific factory is 1 out of 11, and the probability of selecting a specific factory from the remaining selection is 1 out of 10. Therefore, the probability of both factories with the same owner being selected is:

(1/11) * (1/10) = 1/110

So, the probability that both factories with the same owner will be included in the inspection is 1 in 110.