Respuesta :
Answer:
0.2097 = 20.97% probability that your longest wait is less than 12 minutes
Step-by-step explanation:
To solve this question, we need to understand the uniform and the binomial distribution.
Uniform distribution:
A distribution is called uniform if each outcome has the same probability of happening.
The uniform distribution has two bounds, a and b, and the probability of finding a value lower than x is given by:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
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.
Probability of a single bus having a waiting time of less than 12 times:
Uniformly distributed on the interval [0, 15] means that [tex]a = 0, b = 15[/tex]
[tex]P(X < 12) = \frac{12 - 0}{15 - 0} = 0.8[/tex]
What is the probability that your longest wait is less than 12 minutes?
This is the probability that all 7 buses have waiting time less than 12 minutes, which is [tex]P(X = 7)[/tex] when [tex]n = 7[/tex], with [tex]p = 0.8[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{7,7}.(0.8)^{7}.(0.2)^{0} = 0.2097[/tex]
0.2097 = 20.97% probability that your longest wait is less than 12 minutes
