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.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
In this problem we have that:
There are 4 machines, so
.
The probability that on a given day any one machine will break down is 0.53. This means that
.
What is the probability that at least 2 machines will break down on a given day?
![P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
In which
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 2) = C_{4,2}.(0.53)^{2}.(0.47)^{2} = 0.3723](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B4%2C2%7D.%280.53%29%5E%7B2%7D.%280.47%29%5E%7B2%7D%20%3D%200.3723)
![P(X = 3) = C_{4,3}.(0.53)^{3}.(0.47)^{1} = 0.2799](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B4%2C3%7D.%280.53%29%5E%7B3%7D.%280.47%29%5E%7B1%7D%20%3D%200.2799)
![P(X = 4) = C_{4,4}.(0.53)^{4}.(0.47)^{0} = 0.0789](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B4%2C4%7D.%280.53%29%5E%7B4%7D.%280.47%29%5E%7B0%7D%20%3D%200.0789)
So
![P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.3723 + 0.2799 + 0.0789 = 0.7311](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.3723%20%2B%200.2799%20%2B%200.0789%20%3D%200.7311)
There is a 73.11% probability that at least 2 machines will break down on a given day.