Question

William uses a DVD rental service to satisfy his movie needs. It takes 4 days for William to receive a new movie. William's daily thirst for movies is Poisson-distributed with mean 0.5. If William is unable to watch the movies he desires because they are not available, then he watches them as soon as they become available. What is the probability that, on a given day, he cannot satisfy his movie-watching needs?

a. About 15%

b. Cant decide

c. About 2%

d. About 5%

Answer 1

The probability that William cannot satisfy his movie-watching needs on a given day is approximately 0.6065, which is around 61%.

Step-by-step explanation:

The given scenario follows a Poisson distribution with a mean of 0.5 movies per day.

The probability that William cannot satisfy his movie-watching needs on a given day can be found using the Poisson distribution formula.

Let's denote the random variable X as the number of movies William watches on a given day. In this case, we want to find the probability that X is equal to 0, meaning William cannot satisfy his movie-watching needs.

The formula for the Poisson distribution is:

P(X = k) = (e^(-λ) * λ^k) / k!

Where λ is the mean of the distribution and k is the number of occurrences we are interested in.

In this scenario, λ is equal to 0.5 and k is equal to 0.

Plugging these values into the formula:

P(X = 0) = (e^(-0.5) * 0.5^0) / 0! = e^(-0.5) ≈ 0.6065

Therefore, the probability that, on a given day, William cannot satisfy his movie-watching needs is approximately 0.6065, which is around 61%.