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hellppp s1...cannot do this! :(

deededeeddeed.pnganyone?:frown:
Damn and I thought I was getting better at probability. Anyone care to help me on this as well.
Ive tried rearranging the conditional probability formula thing but don't seem to be getting anywhere
Reply 2
This looks like a brutal question. I'll give it a shot in the morning, but is this a Solomon question or something?
Original post by Mr Tall
deededeeddeed.pnganyone?:frown:

Original post by Super199
Damn and I thought I was getting better at probability. Anyone care to help me on this as well.
Ive tried rearranging the conditional probability formula thing but don't seem to be getting anywhere


I've got to the answer, but that took me about 20 minutes at least! That is one bloody difficult question, I'll tell you that. We're not allowed to give answers here, but I'll give you a few clues that helped me.

Bear in mind these facts:

P(A')=1-P(A)
P(B∩A)+P(B∩A')=P(B)

They've given you three pieces of information, so write them out in full mathematically. The first thing I did was work out P(A) in terms of P(B). Then, try to substitute your equations together and you should find that you'll be able to get everything in terms of P(A), P(B) and P(B
∩A), and then use your substitutions to get it all in terms of P(A).
(edited 9 years ago)
Reply 4
P(B | A) + P(B | A') = P(B) (I think :redface:)

P(B) = 1/4 + 1/8 = 3/8

P(A | B) = P(A intersection B) / P(B)

P(A intersection B) = 1/5 x 3/8 = 3/40

P(A intersection B) = P(B intersection A)

P(B | A) = P(B intersection A) / P(A)

P(A) x P(B | A) = P(B intersection A)

P(A) = P(B intersection A) / P(B | A) = 3 / 40 x 4 / 1 = 12 / 40

I know I'm wrong, but I can't think of any other way of doing it. Hopefully, this working isn't far off (it's 11pm and I'm shattered), and may prove useful in working out the answer.
Original post by VannR
P(B | A) + P(B | A') = P(B) (I think :redface:)

P(B) = 1/4 + 1/8 = 3/8

P(A | B) = P(A intersection B) / P(B)

P(A intersection B) = 1/5 x 3/8 = 3/40

P(A intersection B) = P(B intersection A)

P(B | A) = P(B intersection A) / P(A)

P(A) x P(B | A) = P(B intersection A)

P(A) = P(B intersection A) / P(B | A) = 3 / 40 x 4 / 1 = 12 / 40

I know I'm wrong, but I can't think of any other way of doing it. Hopefully, this working isn't far off (it's 11pm and I'm shattered), and may prove useful in working out the answer.


Your first step isn't correct, unfortunately. P(B∩A)+P(B∩A')=P(B), not what you've written.

Just a word of warning to everyone: if you can't do this, don't worry. This is a million times more difficult than anything that could come up in an S1 paper.
(edited 9 years ago)
Reply 6
Use the other equation given on the formula sheet for S1 probability.

P(A|B) = (P(B|A) * P(A)) / (P(B|A)*P(A)+P(B|notA)*P(not A))

Then use 1-P(A) = P(notA)
Reply 7
Original post by Mr Tall
deededeeddeed.pnganyone?:frown:


This is just a case of writing down every formula you know and then ploughing through the algebra.

Start by writing down 2 different ways of calculating P(A n B):

P(A n B) = P(A|B) x P(B) = (1/5)P(B)
P(A n B) = P(B|A) x P(A) = (1/4)P(A)

so P(B) = (5/4)P(A)

Also, P(B) = P(B n A) + P(B n A') since B can either happen with A or without it, but not both.

And P(B n A') = P(B|A') x P(A') = (1/8) x P(A') = (1/8) x (1 - P(A))

Now you can put all the bits together and finish it off :smile:

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