Respuesta :
Answer:
59.41% probability that at least 1 was not on time.
Step-by-step explanation:
For each arrival, there are only two possible outcomes. Either it is on time, or it is not. The probability of an arrival being on time is independent from other arrivals, so we use the binomial probability distribution to solve this problem.
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.
In this problem we have that:
[tex]n = 5, p = 0.835[/tex]
Find the probability that at least 1 was not on time.
Either all of them were on time, or at least one of them was late. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 5) + P(X \leq 4) = 1[/tex]
We want to find [tex]P(X \leq 4)[/tex]. So
[tex]P(X \leq 4) = 1 - P(X = 5)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.835)^{5}.(0.165)^{0} = 0.4059[/tex]
[tex]P(X \leq 4) = 1 - P(X = 5) = 1 - 0.4059 = 0.5941[/tex]
59.41% probability that at least 1 was not on time.
