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:There is a 73.11% probability that at least 2 machines will break down on a given day.Step-by-step explanation:Binomial probability distributionThe 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.