Respuesta :
Answer:
There is a 73.11% probability that at least 2 machines will break down on a given day.
Step-by-step explanation:
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:
There are 4 machines, so [tex]n = 4[/tex].
The probability that on a given day any one machine will break down is 0.53. This means that [tex]p = 0.53[/tex].
What is the probability that at least 2 machines will break down on a given day?
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{4,2}.(0.53)^{2}.(0.47)^{2} = 0.3723[/tex]
[tex]P(X = 3) = C_{4,3}.(0.53)^{3}.(0.47)^{1} = 0.2799[/tex]
[tex]P(X = 4) = C_{4,4}.(0.53)^{4}.(0.47)^{0} = 0.0789[/tex]
So
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.3723 + 0.2799 + 0.0789 = 0.7311[/tex]
There is a 73.11% probability that at least 2 machines will break down on a given day.
