conditional probability of circuit fault

[Rose_01298]

A test procedure for diagnosing faults in circuits indicates no fault with probability 0.99 if the circuit is faultless. It indicates a fault with probability 0.90 if the circuit is faulty. Let the probability of a circuit to be faulty be 0.02.
(1) What is the probability that a circuit is faulty if the test procedure indicates a fault?

(2) What is the probability that a circuit is faultless if the test procedure indicates that it is faultless?

 

[Deleted member 4993]

A test procedure for diagnosing faults in circuits indicates no fault with probability 0.99 if the circuit is faultless. It indicates a fault with probability 0.90 if the circuit is faulty. Let the probability of a circuit to be faulty be 0.02.
(1) What is the probability that a circuit is faulty if the test procedure indicates a fault?

(2) What is the probability that a circuit is faultless if the test procedure indicates that it is faultless?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

 

[Rose_01298]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

This is what I have so far.
P(no fault|faultless)=0.991
P(fault|faulty)=0.90
P(faulty)=0.2
P(faultless)=0.80
This is how I interpreted the given information in the problem.
I assume (1) is asking P(fault|faulty) and (2) is asking P(no fault|faultless). Is this accurate?
Because if so then I think the answer for (1) would be 0.90 and (2) would be 0.99. But that is where I am stuck. I think I am interpreting the information wrong.

 

[Steven G]

The 1st probability is not correct--probably just a typo.
The 3rd probability you listed is wrong--just read the given information a bit more carefully.
The 4th probability is wrong because you had the wrong 3rd probability.
Fix these mistakes and work out your answer again.

 

[Rose_01298]

The 1st probability is not correct--probably just a typo.
The 3rd probability you listed is wrong--just read the given information a bit more carefully.
The 4th probability is wrong because you had the wrong 3rd probability.
Fix these mistakes and work out your answer again.

With the revisions, would I write it like this?
P(no fault|not a faulty circuit)=0.99
P(fault|faulty circuit)=0.90
P(faulty circuit)=0.02
P(not a faulty circuit)=0.98

And then part 1 is asking P(faulty circuit|fault)
And part 2 is asking P(not a faulty circuit| no fault)?

 

[pka]

With the revisions, would I write it like this?
P(no fault|not a faulty circuit)=0.99
P(fault|faulty circuit)=0.90
P(faulty circuit)=0.02
P(not a faulty circuit)=0.98

And then part 1 is asking P(faulty circuit|fault)
And part 2 is asking P(not a faulty circuit| no fault)?

First, thank you for the above notation, for I simply got lost in the "double nagations".
Part 1. \(\mathcal{P}(\text{faulty circuit}|\text{fault})=\dfrac{\mathcal{P}(\text{faulty circuit}\cap\text{fault})}{\mathcal{P}(\text{fault})}\)\(=\dfrac{\mathcal{P}(\text{fault})|\text{faulty circuit})\mathcal{P}(\text{fault circuit})}{\mathcal{P}(\text{fault})}\)
Can you carry on?

 

[HallsofIvy]

A test procedure for diagnosing faults in circuits indicates no fault with probability 0.99 if the circuit is faultless. It indicates a fault with probability 0.90 if the circuit is faulty. Let the probability of a circuit to be faulty be 0.02.
(1) What is the probability that a circuit is faulty if the test procedure indicates a fault?

(2) What is the probability that a circuit is faultless if the test procedure indicates that it is faultless?

My standard method for such problems:

Imagine 10000 such circuits. 0.02(10000)= 200 of them have a fault, 980 do not.

Of the 200 that have a fault, .99(200)= 198 test faulty, 2 do not.
Of the 980 without a fault, .10(980)= 98 test faulty, 898 do not.

So we have a total of 198+ 98= 296 that test faulty of which 198 really are faulty. That is \(\displaystyle \frac{198}{296}= 0.6689...\) or 67%.

And we have a total of 2+ 898= 900 that do not test faulty of which 898 do not have a fault. That is \(\displaystyle \frac{898}{900}= 0.99778\) or 100%.

 

[Dr.Peterson]

Imagine 10000 such circuits. 0.02(10000)= 200 of them have a fault, 980 do not.

You meant 9800, right? But the OP can make the corrections.

I like to do essentially the same thing, but in a table; and in checking the rows and columns of the table, I'd catch such errors. Which I do make.