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.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
![f(x) = \mu e^{-\mu x}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cmu%20e%5E%7B-%5Cmu%20x%7D)
In which
is the decay parameter.
The probability that x is lower or equal to a is given by:
![P(X \leq x) = \int\limits^a_0 {f(x)} \, dx](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%20%5Cint%5Climits%5Ea_0%20%7Bf%28x%29%7D%20%5C%2C%20dx)
Which has the following solution:
![P(X \leq x) = 1 - e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%201%20-%20e%5E%7B-%5Cmu%20x%7D)
The probability of finding a value higher than x is:
![P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%3E%20x%29%20%3D%201%20-%20P%28X%20%5Cleq%20x%29%20%3D%201%20-%20%281%20-%20e%5E%7B-%5Cmu%20x%7D%29%20%3D%20e%5E%7B-%5Cmu%20x%7D)
Probability of a single tower being standing after 35 years:
Single tower, so exponential.
Mean of 25 years, so ![m = 25, \mu = \frac{1}{25} = 0.04](https://tex.z-dn.net/?f=m%20%3D%2025%2C%20%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B25%7D%20%3D%200.04)
We have to find ![P(X > 35)](https://tex.z-dn.net/?f=P%28X%20%3E%2035%29)
![P(X > 35) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-0.04*35} = 0.2466](https://tex.z-dn.net/?f=P%28X%20%3E%2035%29%20%3D%201%20-%20P%28X%20%5Cleq%20x%29%20%3D%201%20-%20%281%20-%20e%5E%7B-%5Cmu%20x%7D%29%20%3D%20e%5E%7B-0.04%2A35%7D%20%3D%200.2466)
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 ![p = 0.2466](https://tex.z-dn.net/?f=p%20%3D%200.2466)
3 towers, so ![n = 3](https://tex.z-dn.net/?f=n%20%3D%203)
We have to find:
![P(X \geq 2) = P(X = 2) + P(X = 3)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29)
In which
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 2) = C_{3,2}.(0.2466)^{2}.(0.7534)^{1} = 0.1374](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B3%2C2%7D.%280.2466%29%5E%7B2%7D.%280.7534%29%5E%7B1%7D%20%3D%200.1374)
![P(X = 3) = C_{3,3}.(0.2466)^{3}.(0.7534)^{0} = 0.0150](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B3%2C3%7D.%280.2466%29%5E%7B3%7D.%280.7534%29%5E%7B0%7D%20%3D%200.0150)
![P(X \geq 2) = P(X = 2) + P(X = 3) = 0.1374 + 0.0150 = 0.1524](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%3D%200.1374%20%2B%200.0150%20%3D%200.1524)
15.24% probability that at least 2 will still stand after 35 years