Okay here I am again with problems that I can't solve within the time that my patience lasts... I find them quite hard to solve.

It's actually my goal to get somekind of approach to solve these problems, since the answer is in the back of the book... But I can't find a way to get to the answer, which annoys me

QUESTION 1

Consider the diagram of lights below. Each lamp has a 99% chance of producing light. What is the probability that light will appear out of this system?



(Correct answer: 99,9997980399%)

QUESTION 2

Consider this diagram of lights below. Each lamp can break independently of the others. The chances of failing are given in the table.

Given there is no connection between starting and ending point, what is the probability that the failure lies with A?



(Correct answer: 73,53%)

QUESTION 3

A disease affects 6,5% of the population. There is however an inheritance factor. If one's father has the disease, the probabilty that one gets the disease is 13%. What is the probability of getting the disease when the father never had the disease?

(there is no info missing here)

(Correct answer: 6,05%)

THX! A golden star for the people who can solve these problems.

In order to get this answer, I must assume that AF alone and CD alone cannot produce light. That is, there is no path out of A and into F, and no path out of C and into D, even though there appears to be one the way it is drawn, and in the real world there would be such a connection at the output of B. If these paths worked, the answer would be 99.9998000001%.

To have light, we must have a complete circuit from one pole of the battery to the other through working lights, since a failed light will break its part of the circuit. We need at least 1 of {A,B,C} and at least 1 of {D,E,F}, but AF alone and CD alone are not sufficient.

Let A,B,C mean that these are working, and A', B', C' mean that these are not working. P(ABC) is the probability that A,B, and C are working, and P(A'B'C') means that A, B, and C are not working.

P(light) = 1 - P(no light)

= 1 -
[ P(A'B'C') + P(D'E'F') - P(A'B'C'D'E'F') +
P(AFB'C'D') + P(CDA'B'E'F') ]

Note that P(A'B'C') + P(D'E'F') counts P(A'B'C'D'E'F') twice, so we subtract this off (inclusion-exclusion principle).

Now P(A'B'C') = P(D'E'F') = (0.01)³,
P(A'B'C'D'E'F') = (0.01)⁶, and
P(AFB'C'D') = P(CDA'B'E'F') = (0.99)²*(0.01)⁴

P(light) =

1 - [2*(0.01)³ - (0.01)⁶ + 2*(0.99)²*(0.01)⁴]

= 99.9997980399%



P(A' | no light) = P(A' and no light) / P(no light)

= P(A'D'E') / [ P(A'D'E') + P(AB'C'D'E') ]

= (0.1)*(0.3)*(0.4) / [ (0.1)*(0.3)*(0.4) + (0.9)(0.2)(0.2)(0.3)(0.4) ]

=~ 73.53%


Let D = disease
F = father had disease
F' = father did not have disease
want P(D | F')

P(D) = P(D and F) + P(D and F')

P(D) = P(D | F)*P(F) + P(D | F')*P(F')

0.065 = (0.13)*(0.065) + P(D | F')*(1 - 0.065)

P(D | F') = [ 0.065 - (0.13)*(0.065) ] / (1 - 0.065)

P(D | F') =~ 6.05%


EDIT: Fixed typos noted by Riverdale.

You make it all sound so easy... You are indeed somebody with skills in this matter.

Can I ask you how you gathered these skills? In your education? Or ... ?

With all due respect, I think you made a mistake in some parts of the first exercise... At least a mistake in notation I mean then... I will put quotes around the things I think are noted down wrong.

You say for example:

Another:

And yet another:

What do you think about this? If I'm all wrong, I didn't understand what you did... But since the P(...) notations don't match up with the powers you used in the calculations, maybe I could be right about this?

I hope to God I'm right otherwise I still didn't understand it...

Thanks! And also big thanks for the other exercises!

Yes, these are all typos. I will fix the original post.

As for me, I'm an electrical engineer with degrees in math and physics also. I specialized in DSP and telecommunications which are heavily probability oriented, and probability was a concentration for my math degree. I've been doing problems here for about 10 years now, which helps alot to keep me sharp. Thanks for the kind words.

bruce, can you expain this in a little more depth:

the last two arguments in P(no light) are throwing me off and i think it's b/c i'm not reading the circuit properly

Well he assumes that because otherwise he wouldn't get the answer that I postulated. If there is a connection between A and F or C and D, then just one of the lights on the left, and one of the lights on the right is enough to get light in this system.

But since there is no such connection, if all lamps but A and F are broken, there still is no light, since there is no connection. The same holds for C and D.

i got that part, i was just reading the circuit incorrectly before

Aha okay