Probability and Subcontractors

[TheClincher]

Homework Statement

A contractor has two subcontractors for his excavation work. Experience shows that in 60% of the time, subcontractor A was available to do a job, whereas subcontractor B was available 80% of the time. Also, the contractor is able to get at least one of these two subcontractors 90% of the time.
a) Probability that both subcontractors will be available to do the next job.
b) If contractor learned that subcontractor A is not available for the job, what is the probability that the other subcontractor will be available?

Homework Equations

A = sub A available
B = sub B available

The Attempt at a Solution

a) P(AB) = P(A)P(B)

b is the one I have a question on. I think it's the following set up but I don't know how to continue or what to do with it:
b) P(B|\bar{A}) = P(B\bar{A})/P(\bar{A})

Okay uhh I don't know if this is sensible but I took P(B\bar{A}) and did the following

P(B\bar{A}) = 1-complement of P(B\bar{A}) and expanded it to get 0.32. So if I plug that into P(B|\bar{A}) = P(B\bar{A})/P(\bar{A}) I got 0.32/0.40 = 0.80. Eh? Can someone steer me right

 

[Dick]

I think you are making a mistake in assuming that the availability of the two contractors is independent. This makes P(AB)=P(A)P(B) wrong. Draw a Venn diagram type thing for the availability space. P(A) is 0.6, P(B) is 0.8 and the region outside of P(A) or P(B) is 0.1. Reason from there. What's the correct value for P(AB)?

 

[TheClincher]

Oh, so, hmm... P(AB) = P(A) + P(B) - P(A U B) = 0.50?

 

[Dick]

Oh, so, hmm... P(AB) = P(A) + P(B) - P(A U B) = 0.50?

That's what I get. Now how about the second part?

 

[TheClincher]

(I'll denote complements like this: c{A} for complement of A)

B is continuing to vex me. From the problem statement, I'm interpreting this: P(c{A}B U Ac{B}) = P(c{A}B) + P(Ac{B}) = 0.90; that is, if one is not available the contractor still has a 90% chance of being able to hire the other subcontractor. I think I have to expand the two elements P(c{A}B) + P(Ac{B}), but I don't think I know how. :-\

 

[Dick]

Try drawing a picture first. It really helps. Draw a circle A which represents .6, a circle B that represents .8, outside of them both represents .1. From a) we know the overlap between A and B is .5. That leaves .1 in A outside of B and .3 in B outside of A. Now b) is easy. The total area outside of A is .4, B covers .3 of that. So the odds are? Now try and translate that picture into your set notation.

 

[TheClincher]

ah well I've got the solution and the concept. drawing venn diagrams certainly does help. i spent last night focusing on a purely graphical approach instead of a symbolic-computation approach on a different probability problem and solved it successfully. thank you for your guidance, Dick.

 

[helpme74]

So I have drawn the venn diagram and I understand what you are saying about the areas, but I still don't quite understand what area constitutes my answer. I want to assume that the two, A and B, are statistically independent, which would give .8, but i know that's too easy. A little more help please...

 

[Dick]

The area you want for a) is the area of the intersection of A and B. Let O be the area of the intersection of A and B. Now solve for all of the areas. DON'T assume they are statistically independent. Even if they were, you still wouldn't get .8.

 

[helpme74]

P(AUB) = .9, P(AB) = .5; the compliment of A is .4, if the compliment of A occurs, that leaves .6 insides the sample space, right? With .1 of that being the intersection of compliment of A and compliment of B, so that leaves .5 left, but that doesn't make sense looking at the venn diagram.
Am I still taking the wrong approach?

 

[Dick]

P(AUB) = .9, P(AB) = .5; the compliment of A is .4, if the compliment of A occurs, that leaves .6 insides the sample space, right? With .1 of that being the intersection of compliment of A and compliment of B, so that leaves .5 left, but that doesn't make sense looking at the venn diagram.
Am I still taking the wrong approach?

Sorry, I should have said intersection instead of union in my last post. I corrected it. And, yes, P(AB)=.5. My venn diagram makes sense P(A)=.6=.5+.1, P(B)=.8=.3+.5. What part of yours doesn't make sense?

 

[helpme74]

I mean when looking at the diagram: if the compliment of A occurs, that means that A will not occur, and P(AB) will also not occur; that just leaves .3 left inside of B, and I'm getting .5 for the answer to the question, not .3

 

[Dick]

I mean when looking at the diagram: if the compliment of A occurs, that means that A will not occur, and P(AB) will also not occur; that just leaves .3 left inside of B, and I'm getting .5 for the answer to the question, not .3

Which question? Are you trying answer b) or a)?

 

[helpme74]

Please forgive me for not clarifying earlier, I'm trying to answer (b). I know that .5 is the answer for (a), but I'm getting .5 and .3 for answers for (b).

I got .5 by going this route:
P(AUB) = .9, P(AB) = .5; the compliment of A is .4, if the compliment of A occurs, that leaves .6 insides the sample space, right? With .1 of that being the intersection of compliment of A and compliment of B, so that leaves .5 left for my answer.

And I got .3 by looking at the diagram and trying to solve visually:
When looking at the diagram: if the compliment of A occurs, that means that A will not occur, and P(AB) will also not occur; that just leaves .3 left inside of B for my answer.

This seems rather simple but I am still confused.

 

[Dick]

Please forgive me for not clarifying earlier, I'm trying to answer (b). I know that .5 is the answer for (a), but I'm getting .5 and .3 for answers for (b).

I got .5 by going this route:
P(AUB) = .9, P(AB) = .5; the compliment of A is .4, if the compliment of A occurs, that leaves .6 insides the sample space, right? With .1 of that being the intersection of compliment of A and compliment of B, so that leaves .5 left for my answer.

And I got .3 by looking at the diagram and trying to solve visually:
When looking at the diagram: if the compliment of A occurs, that means that A will not occur, and P(AB) will also not occur; that just leaves .3 left inside of B for my answer.

This seems rather simple but I am still confused.

If you are trying to do b) then you exclude the region where A is available. That leaves you with an area of .3 in B and and area of .1 outside of B, right? What's the PROBABILITY (not area) that B is available?

 

[helpme74]

Ah, yes. I think perhaps I 'over-complicated' things.
Thank you VERY much for your assistance, sir. I am indebted.