Answer:
0.99 probability that at least one trip will last less than an hour
Step-by-step explanation:
For each time that the machine is used, there are only two possible outcomes. Either it lasts more than an hour, or it does not. The probability of a trip lasting more than an hour is independent of other trips. So the binomial probability distribution is used to solve this question.
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 combinations 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.
0.8 probability of lasting more than one hour.
So [tex]p = 0.8[/tex]
20 trips
So [tex]n = 20[/tex]
Assuming that each trip is equally likely to last for more than an hour, what is the probability that at least one trip will last less than an hour?
Either all the trips last for more than an hour, or at least one does not. The sum of the probabilities of these outcomes is decimal 1. So
[tex]P(X = 20) + P(X < 20) = 1[/tex]
We want P(X < 20). So
[tex]P(X < 20) = 1 - P(X = 20)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 20) = C_{20,20}.(0.8)^{20}.(0.2)^{0} = 0.0115[/tex]
[tex]P(X < 20) = 1 - P(X = 20) = 1 - 0.0115 = 0.9885[/tex]
0.9885 probability that at least one trip will last less than an hour
Rouding to the nearest hundreth.
0.99 probability that at least one trip will last less than an hour
