Respuesta :
Answer:
a) There is a 99.99997% probability that at least one valve opens.
b) There is a 22.62% probability that at least one valve fails to open.
Step-by-step explanation:
For each valve, there are only two possible outcomes. Either it open, or it does not. So 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 combinations 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]p = 0.95, n = 5[/tex]
(a) What is the probability that at least one valve opens?
Either no valves open, or at least one opens. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want [tex]P(X \geq 1)[/tex]
So
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
[tex]P(X = 0) = C_{5,0}.(0.95)^{0}.(0.05)^{5} = 0.0000003[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000003 = 0.9999997[/tex]
There is a 99.99997% probability that at least one valve opens.
(b) What is the probability that at least one valve fails to open?
Either all valves open, or at least one does not open. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 5) + P(X \leq 4) = 1[/tex]
We want [tex]P(X \leq 4)[/tex]
So
[tex]P(X \leq 4) = 1 - P(X = 5)[/tex]
[tex]P(X = 5) = C_{5,5}.(0.95)^{5}.(0.05)^{0} = 0.7738[/tex]
[tex]P(X \leq 4) = 1 - P(X = 5) = 1 - 0.7738 = 0.2262[/tex]
There is a 22.62% probability that at least one valve fails to open.
