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

Answer:

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.

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.