Question

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given $Table4.37.$There is five-video limit per customer at this store, so nobody ever rents more than five DVDs.

a. Describe the random variable X in words.

b. Find the probability that a customer rents three DVDs.

c. Find the probability that a customer rents at least four DVDs.

d. Find the probability that a customer rents at most two DVDs.

Short answer

a. $"x"$is the number of DVDs a customer rents

b. Probability that a customer rents three DVDs are $0.11$

c. Probability that a customer rents at least four DVDs are $0.09$

d. Probability that a customer rents at most two DVDs are$0.80$

Step-by-step solution

Given information

The probability distribution for DVD rentals per customer at Video To Go is given $Table4.37$

Part (a) Step 1: Explanation

Describe the random variable $X$in words.

1. $X$is the number of DVDs a customer owns
2. $X$ is the total number of DVDs rented at the store
3. $X$ is the number of DVDs a customer rents
4. $X$ is the number of customers that come into the store
5. $X$ is the number of times a single customer comes into the store

Part (b) Step 1: Explanation

Since the probability is$100\%$that a person who comes in the store will rent either $0,1,2,3,4,or5$DVDs, then we just subtract the others from $100\%$to get the probability they will rent $3$DVDs.

So$100\%-4\%-52\%-24\%-6\%-3\%=11\%$. Therefore, the probability that a person who comes in the store will rent three DVDs is $11\%or0.11.$

Part (c) Step 1: Explanation

The probability that a customer rents at least four DVDs is equal to the probability they will rent $4OR5$ DVDs so $6\%+3\%=9\%or0.09.$

Part (d) Step 1: Explanation

The probability that a customer rents at most two DVDs is equal to te probability they will rent zero OR one, OR two DVDs so$4\%+52\%+24\%=80\%or0.80.$