Answer:
There is a 99.99998% probability that at least one valve opens.
Step-by-step explanation:
For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 5, p = 0.93[/tex]
Calculate P(at least one valve opens).
This is [tex]P(X \geq 1)[/tex]
Either no valves open, or at least one does. The sum of the probabilities of these events is decimal 1. So:
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.93)^{0}.(0.07)^{5} = 0.0000016807[/tex]
Finally
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000016807 = 0.9999983193[/tex]
There is a 99.99998% probability that at least one valve opens.
