Respuesta :
Answer:
15.24% probability that at least 2 will still stand after 35 years
Step-by-step explanation:
To solve this question, we need to understand the binomial distribution and the exponential distribution.
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.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
Probability of a single tower being standing after 35 years:
Single tower, so exponential.
Mean of 25 years, so [tex]m = 25, \mu = \frac{1}{25} = 0.04[/tex]
We have to find [tex]P(X > 35)[/tex]
[tex]P(X > 35) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-0.04*35} = 0.2466[/tex]
What is the probability that at least 2 will still stand after 35 years?
Now binomial.
Each tower has a 0.2466 probability of being standing after 35 years, so [tex]p = 0.2466[/tex]
3 towers, so [tex]n = 3[/tex]
We have to find:
[tex]P(X \geq 2) = P(X = 2) + P(X = 3)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.2466)^{2}.(0.7534)^{1} = 0.1374[/tex]
[tex]P(X = 3) = C_{3,3}.(0.2466)^{3}.(0.7534)^{0} = 0.0150[/tex]
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) = 0.1374 + 0.0150 = 0.1524[/tex]
15.24% probability that at least 2 will still stand after 35 years
