Question

Solar-heat installations successfully reduce the utility bill 60% of the time. What is the probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill?

Answer 1

Answer:

4.23% probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill.

Step-by-step explanation:

For each installation, there are only two possible outcomes. Either it reduces the utility bill, or it does not. The probabilities for each installation reducing the utility bill are independent. So we use the binomial probability distribution to solve this question.

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 combinations 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:

Solar-heat installations successfully reduce the utility bill 60% of the time, which means that $p = 0.6$

What is the probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill?

This is $P(X \geq 9)$ when $n = 10$. So

$P(X \geq 9) = P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.6)^{9}.(0.4)^{1} = 0.0363$

$P(X = 10) = C_{10,10}.(0.6)^{10}.(0.4)^{0} = 0.0060$

So

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0363 + 0.0060 = 0.0423$

4.23% probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill.