Question

Consider the following binomial experiment. A company owns 4 machines. The probability that on a given day any one machine will break down is 0.53. What is the probability that at least 2 machines will break down on a given day?a) 0.6960b) 0.5398c) 0.8638d) 0.0225e) 0.5806

Answer 1

There is a 73.11% probability that at least 2 machines will break down on a given day.

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)`

In which
`C_(n,x)` is the number of different combinatios of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

There are 4 machines, so
`n = 4`.

The probability that on a given day any one machine will break down is 0.53. This means that
`p = 0.53`.

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)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 2) = C_(4,2).(0.53)^(2).(0.47)^(2) = 0.3723`

`P(X = 3) = C_(4,3).(0.53)^(3).(0.47)^(1) = 0.2799`

`P(X = 4) = C_(4,4).(0.53)^(4).(0.47)^(0) = 0.0789`

So

`P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.3723 + 0.2799 + 0.0789 = 0.7311`

There is a 73.11% probability that at least 2 machines will break down on a given day.