Question

Consider a series system composed of 4 separate components where each component has a 30% chance of failing. Assume each component functions independent of each other and the system will fail if at least one of the components fails. If you have 7 of these systems, what is the probability that exactly 3 of them would function? Clearly state the events of interest and random variable of interest using the context of the problem. Hint: Find the probabi

Answer 1

Answer:

16.15% probability that exactly 3 of them would function

Step-by-step explanation:

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 each system working:

4 components, which means that $n = 4$

Each has a 30% probability of failing, so $p = 1 - 0.3 = 0.7$

For the system to work, all 4 components have to work. This is P(X = 4).

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

$P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401$

0.2401 probability of a system working.

If you have 7 of these systems, what is the probability that exactly 3 of them would function?

Now 7 systems, so $n = 7$

0.2401 probability of a system working.

We have to find P(X = 3).

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

$P(X = 3) = C_{7,3}.(0.2401)^{3}.(0.7599)^{4} = 0.1615$

16.15% probability that exactly 3 of them would function