Question

f) The life of a power transmission tower is exponentially distributed, with mean life 25 years. If three towers, operated independently, are being erected at the same time, what is the probability that at least 2 will still stand after 35 years?

Answer 1

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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Probability of a single tower being standing after 35 years:

Single tower, so exponential.

Mean of 25 years, so $m = 25, \mu = \frac{1}{25} = 0.04$

We have to find $P(X > 35)$

$P(X > 35) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-0.04*35} = 0.2466$

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 $p = 0.2466$

3 towers, so $n = 3$

We have to find:

$P(X \geq 2) = P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{3,2}.(0.2466)^{2}.(0.7534)^{1} = 0.1374$

$P(X = 3) = C_{3,3}.(0.2466)^{3}.(0.7534)^{0} = 0.0150$

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.1374 + 0.0150 = 0.1524$

15.24% probability that at least 2 will still stand after 35 years