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

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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.