Conditional Probability

[y.v]

I'm having some difficulties with the calculation of conditional probability using the formula:

P (A|B) = (P (A ∩ B)) / (P(B))

For example, how would you calculate:

On a given day, photocopier A has a 10% chance of a malfunction and machine B has a 7% chance of malfunction. Given that atleast one of the machines has malfunctioned, what is the chance that it was machine B?

This is what I tried:

P (Machine A has malfunctioned provided that one of the machines has malfunctioned):

P (A malfunctions) = 7/100

P (One of the machines malfunction) = 17/100 [?]

... I need help from here onwards!

Edited May 4, 2009 by y.v

[Sandwich]

On a given day, photocopier A has a 10% chance of a malfunction and machine B has a 7% chance of malfunction. Given that at least one of the machines has malfunctioned, what is the chance that it was machine B?

Okay, so

P(A) = 0.1

P(B) = 0.07

Probability of either of the machines malfunctioning = 0.17 (because they are independent, so the chances of one malfunctioning don't affect the other)

Then the probability of B was 0.07, so of the 0.17 it makes up 0.07 of it all .:. P of B given that one of the machines has malfunctioned ought to be 0.07/0.17 = 7/17.

I don't think you can use the formula as there is no overlap between the two...? So there's no (A ∩ B).

I hope that's right, anyway. D'you have the answers?

[y.v]

Thanks for answering!

However, the answer is supposed to be: 70/163

[Sandwich]

LOL, so much for that, then ; Sorry about that!

EDIT: I keep trying and I really can't get the answer. If you do work it out, mind posting it up? I am intrigued now

Edited May 4, 2009 by Sandwich

[diabolicalangle]

i totally would have agreed wit sandwich... cept its wrong?

i swear that was how it was done. soo frustrating

[moneyfaery]

I've always liked ugly tree diagrams...

Prob of either breaking = 0.10(0.93) + 0.10(0.07) + 0.09(0.07) = 163/1000

Prob of B breaking = 0.10(0.07) + 0.09(0.07) = 70/1000

Prob of B breaking given something broke = (70/1000)/(163/1000) = 70/163

Try to do this without a calculator as the numbers are pretty nice. I converted it to decimal format because it's easier to show in a forum, haha.

Btw, for indep events, you never add the probabilities. You always multiply.

Edited May 4, 2009 by Irene

[Max]

On a given day, photocopier A has a 10% chance of a malfunction and machine B has a 7% chance of malfunction. Given that at least one of the machines has malfunctioned, what is the chance that it was machine B?

So basically:

P(Machine B) = 0.07

P(At least one of the machines has malfunction) = 1 - P(Both machines work) = 1 - 0.93 * 0.90 = 1 - 0.837 = 0.163

--> P(It was machine B|At least one of the machines has malfunction) = P(It was machine B)/P(At least one of the machines has malfunction) = 0.07 / 0.163 = 70/163

I'm afraid I don't know how to use that formula you mentioned in this situation though.

EDIT: Irene beat me

Edited May 4, 2009 by Max

[y.v]

Thank you so much Irene & Max