Question

The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. What is the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week? (Round your answer to four decimal places.)

P(X greaterthanorequalto 5) = __.

Answer 1

There is a 75.65% probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week.

Explanation:

For each teenager, there are only two possible outcomes. Either they watched a rented video at least once during a week, or they did not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinatios of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. This means that
`p = 0.75`.

What is the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week?

Group of 7, so
`n = 7`.

`P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)`.

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 5) = C_(7,5).(0.75)^(5).(0.25)^(2) = 0.3115`

`P(X = 6) = C_(7,6).(0.75)^(6).(0.25)^(1) = 0.3115`

`P(X = 7) = C_(7,7).(0.75)^(7).(0.25)^(0) = 0.1335`

So

`P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) = 0.3115 + 0.3115 + 0.1335 = 0.7565`.

There is a 75.65% probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week.