Question

In Leakwater township, there are two plumbers. On a particular day three Leakwater residents call village plumbers independently of each other. Each resident randomly chooses one of the two plumbers. What is the probability that all three residents will choose the same plumbe

Answer 1

Answer:

25% probability that all three residents will choose the same plumbe

Step-by-step explanation:

For each resident, there are only two possible outcomes. Either they choose the first plumbe, or they choose the second. The plumbers are chosen independent of each other. 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.

Each resident randomly chooses one of the two plumbers.

This means that $p = \frac{1}{2} = 0.5$

Three residents:

This means that $n = 3$

What is the probability that all three residents will choose the same plumb

This is P(X = 0)(all choose the second pumble) or P(X = 3)(all choose the first pumble). So

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

$P(X = 0) = C_{3,0}.(0.5)^{0}.(0.5)^{3} = 0.125$

$P(X = 3) = C_{3,3}.(0.5)^{3}.(0.5)^{0} = 0.125$

$p = P(X = 0) + P(X = 3) = 2*0.125 = 0.25$

25% probability that all three residents will choose the same plumbe