Exercise 12 Page P19

We are finding the probability of choosing two action DVDs followed by a comedy DVD. Since we do not put the DVD back on the shelf before we pick the next one, the picks are not independent. We can use the Probability of Dependent Events rule to find the probability. The probability

P (2 action, then comedy)

is the product of the following three probabilities:

P (1 st action) P (2 nd action ∣ 1 st action) P (3 rd comedy ∣ 1 st action and 2 nd action)

There are

8 + 3 + 5 = 16

DVDs on the shelf, and

8

of these are action DVDs. We select at random, so the DVDs are equally likely to be picked. Hence, the probability of picking an action DVD first is the ratio of the

number of action DVDs

to the

total number of DVDs .

P (1 st action) = \frac{8}{1 6} = \frac{1}{2}

This action DVD is not put back on the shelf, so before the next pick there are

16 - 1 = 15

DVDs on the shelf, and

8 - 1 = 7

of these are action DVDs. We need to use these new numbers to calculate the probability of the second pick.

P (2 nd action ∣ 1 st action) = \frac{7}{1 5}

This second DVD is not put back on the shelf either, so before the third pick there are

15 - 1 = 14

DVDs left on the shelf. Since none of the first two picks are comedy DVDs, there are still the original

3

comedy DVDs on the shelf. We need to use these numbers to calculate the probability of the third pick.

P (3 rd comedy ∣ 1 st action and 2 nd action) = \frac{3}{1 4}

The answer to the question is the product of the three probabilities we found.

P (2 action, then comedy) = \frac{1}{2} \cdot \frac{7}{1 5} \cdot \frac{3}{1 4} = \frac{1}{2 0}

The probability of picking an action DVD twice in a row and then a comedy DVD without replacement is

\frac{1}{2 0} = 0.05 .