# Conditional Probability Problem

• BigFlorida
In summary, if the entire electrical system fails, the probability that component C does not work is 1-P(C).
BigFlorida
An electrical system consists of four components. The system works if components A and B work and either of the components C or D works. The reliability (probability of working) of each component is given by P(A) = 0.9, P(B) = 0.9, P(C) = 0.8, and P(D) = 0.8. Find the probability that (a) the entire system fails to work and (b) component C does not work, given that the entire system fails to work. Assume that the four components work independently.

Thus far I have solved part a by solving for the probability that the entire system works, denoted P(E) where E is the event that the entire system works, then subtracting that from 1; this gave me the result P(E') = 0.2224 (where E' is the complement of E). I am just having trouble solving part b.

I have found that P(C'|E') = [P(C')xP(E'|C')]/P(E') = P(C' n E')/P(E') where n is read "intersect". I really do not see where to go from here though. I have values for P(C') and P(E') but no values for P(E'|C') or P(C' n E'). Any help would be very much appreciated. Thank you in advance.

Which components have to (not) work if C works and the whole system does not work? What does that imply for P(C' n E')?
The other unknown values can be analyzed in the same way.

BigFlorida said:
I have found that P(C'|E') = [P(C')xP(E'|C')]/P(E') = P(C' n E')/P(E') where n is read "intersect". I really do not see where to go from here though. I have values for P(C') and P(E') but no values for P(E'|C') or P(C' n E'). Any help would be very much appreciated. Thank you in advance.

I think [P(C')xP(E'|C')]/P(E') is easier to calculate than P(C' n E')/P(E') and probably this is always true when one uses Bayes' rule. In fact, I think it is probably a general principle to use ##P(A \cap B) = P(A | B) * P(B)## to get rid of any intersections so that it is easier.

Ok, P(E'|C'). We are given that C has failed. There is no C, it is a short circuit. What is the probability of the system failing in this configuration?

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BigFlorida
@verty I could see how that may be slightly easier. So would this be the correct way to go about it then?

[P(C')xP(E'|C')]/P(E') = [P(C')x(1-P(E|C')]/P(E') = [P(C')x(1-A U B U C' U D)]/P(E') = [P(C')x(1-P(A)xP(B)xP(C')xP(D)]/P(E'). As I am new to this type of math, I do not know if the relation I used, P(E'|C') = (1-P(E|C')) would hold true or not.

This is doing my head in, all these symbols. Let's go back to square 1. You have an electrical system, you have components. You are trying to find how vulnerable the system is when C is removed or has failed. It's a real world problem and you are coming up with crazy formulas that don't seem to relate to anything real.

No engineer is going to look at this question and start scribbling all of that junk down, now are they? They are going to say, hmm, P(C'|E'), that seems backwards, I can switch them by using Bayes' rule. Hmm, P(E'|C'), oh but that's obvious, that's how vulnerable the system is when I remove C.

You need to stay focused on what you are trying to do. It should be simple, it should be straight-forward. And you need to think about what is really going on and not what is a bunch of symbol-twisting at the end of the day. This should be an easy question, and I know it's new for you but the truth is usually simple, that's what they always say. So look at this again and try not to make it too complicated, it just isn't worth getting into this forest of symbols when the problem should be an easy one.

Okay? I hope you see now that actually there is an easier way to think about this and most problems can be handled in the same way, the KISS principle.

BigFlorida said:
x(1-A U B U C' U D)/
I assume you mean intersections, not unions, and you need a P() around the event combination.
But why is there a C' In there? This is the expansion of P(E|C').

For the first problem, you already found the probability that A and B are both working. This is a necessary condition for the system to work.
If C is not working, what are the odds that D is working, that is the only way the whole system will work. Remember, these components, A, B, C, and D, are all independent, so you don't need to use conditional formulas, if you don't want to.
1-P(E|C') = P(E'|C') since in this situation (C'), you only have E and E' as possible outcomes.

For part (b), using "brute force engineering" I got an answer of 0.3165. Is this correct?

insightful said:
For part (b), using "brute force engineering" I got an answer of 0.3165. Is this correct?
What brute force are you using? Look back at post #3, given C' means you only have 3 possible functioning parts and all three have to work in order for the system to work.
1- P(A&B&D)= 1-P(A) * P(B) * P(D), since they are independent.

RUber said:
What brute force are you using?
I simply calculated the probabilities for all 13 ways the system can fail and added them to get 0.2224.
Then added together the subset that had C failing to get 0.0704.
Then 0.0704/0.2224 = 0.3165.

Is there another brute force way?

It looks like you found P(C'|E') = P(C' & E') / P(E'). The question asks for P(E'|C') = P(C'&E')/P(C').

RUber said:
It looks like you found P(C'|E') = P(C' & E') / P(E'). The question asks for P(E'|C') = P(C'&E')/P(C').
Does not "(b) component C does not work, given that the entire system fails to work." ask for P(C'|E')? (Sorry, this is the first time I've seen this nomenclature.)

insightful said:
Does not "(b) component C does not work, given that the entire system fails to work." ask for P(C'|E')? (Sorry, this is the first time I've seen this nomenclature.)
Wow...I was reading it backward this whole time. Sorry about that. Yes...your brute force method was correct, and I was doing it wrong.

## What is conditional probability?

Conditional probability is a mathematical concept that measures the likelihood of an event occurring, given that another event has already occurred. It is typically denoted as P(A|B), where A is the event we are interested in and B is the event that has already occurred.

## How is conditional probability calculated?

Conditional probability is calculated by dividing the probability of the joint occurrence of events A and B by the probability of event B. This can be represented as P(A|B) = P(A∩B) / P(B).

## What is the difference between conditional probability and joint probability?

Conditional probability measures the likelihood of one event occurring, given that another event has already occurred. Joint probability, on the other hand, measures the likelihood of two events occurring simultaneously. In other words, conditional probability focuses on a specific scenario while joint probability considers all possible scenarios.

## How is conditional probability used in real-life situations?

Conditional probability is commonly used in various fields such as finance, medicine, and marketing. For example, in finance, conditional probability can be used to calculate the likelihood of a stock's price increasing, given that the overall market has also increased. In medicine, conditional probability can be used to determine the probability of a patient having a certain disease, given their symptoms. In marketing, conditional probability can be used to predict the likelihood of a customer purchasing a product, given their demographic information.

## What are some common misconceptions about conditional probability?

One common misconception about conditional probability is that it can only be used when events are independent. However, conditional probability can also be used for dependent events. Another misconception is that conditional probability can exceed 1, when in fact it is always between 0 and 1. It is important to carefully consider the given information and accurately calculate the probabilities when dealing with conditional probability problems.

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