Question

A k out of n system is one in which there is a group of n components, and the system will function if at least k of the components function. For a certain 4 out of 6 system, assume that on a rainy day each component has probability 0.7 of functioning, and that on a non rainy day each component has probability 0.9 of functioning.Assume that the probability of rain tomorrow is 0.20. What is the probability that the system will function tomorrow

Answer 1

Answer:

0.9606 = 96.06% probability that the system will function tomorrow

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works, or it does not. Components are independent. This means that we use the binomial probability distribution to solve this question.

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.

Probability of a component working:

0.7 of 0.2(rain)

0.9 of 1 - 0.2 = 0.8(no rain). So

$p = 0.7*0.2 + 0.9*0.8 = 0.86$

0.86 - 86% probability that the system will function tomorrow

6 components:

This means that $n = 6$

What is the probability that the system will function tomorrow

This is

$P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 4) = C_{6,4}.(0.86)^{4}.(0.14)^{2} = 0.1608$

$P(X = 5) = C_{6,5}.(0.86)^{5}.(0.14)^{1} = 0.3952$

$P(X = 6) = C_{6,6}.(0.86)^{6}.(0.14)^{0} = 0.4046$

$P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1608 + 0.3952 + 0.4046 = 0.9606$

0.9606 = 96.06% probability that the system will function tomorrow