[pratstercs]

Could someone please run through Conditional Probability (as much as is needed for OCR MEI S1)? I seem to get contradicting information everywhere I look...
For:

I tried

but apparently that's not right. How do I calculate this?
Thanks

[Hopple]

Your second formula is not the same as the first. Think about what conditional probability means, perhaps drawing it out with a trusty Venn diagram.

[pratstercs]

(Original post by Hopple)
Your second formula is not the same as the first. Think about what conditional probability means, perhaps drawing it out with a trusty Venn diagram.

I also try

but then this just equals P(A) and isn't the answer given in the markscheme, so I'm just entirely confused by the whole thing.

[Hopple]

What information have you been given? You need AnB, not A + B. P(A)P(B) would be correct if A and B were independent, though.

[bijesh12]

Conditional Probability is the probability of event A, given that B occurs.

P(A|B) =P(A ∩ B) / P(B)

If P(B) =0, then P(A|B) is undefined.

P(A n B) is the the probablity of the intersection of events A and B. Give a question you're stuck with, and il explain the concept through it

[pratstercs]

(Original post by bijesh12)
Conditional Probability is the probability of event A, given that B occurs.

P(A|B) =P(A ∩ B) / P(B)

If P(B) =0, then P(A|B) is undefined.

P(A n B) is the the probablity of the intersection of events A and B. Give a question you're stuck with, and il explain the concept through it

S1 June 2010
7(iii): Given that the 1200 train is on time, find the probability that the 1000 train is also on time.
The probability of the 1200 train being on time is 0.926375.
The probability of the 1000 train being on time is 0.95.

[bijesh12]

(Original post by pratstercs)
S1 June 2010
7(iii): Given that the 1200 train is on time, find the probability that the 1000 train is also on time.
The probability of the 1200 train being on time is 0.926375.
The probability of the 1000 train being on time is 0.95.

My gut feeling is you need a probablity tree, most probabily from a previous part of the question to answer this. from this tree, you will need to find out the probability of p (A n B)... link me to the paper if you can.

[pratstercs]

I've drawn the tree diagram, that's how I got the probability of the 1200 train, but I don't know how to get P(A∩B) or what to do from there...

Paper (PDF)

[bijesh12]

Let A Be the event where 1000 is on time, when 1200 is on time. This is the conditional probability we are trying to find, denoted P(A |B).

Let B, be the event where 1200 is on time. This is given to us as 0.926375.

You also say P(1000 on time) is 0.95. We only use this if the events A and B are independant (outcome of A is not dependant on B). We have to use this though.

P ( A n B) Is the probablity that both A (1000 on time) and B(1200 is on time). Here we refer to the probability tree to find this out :

P(1000 is on time) = 0.95*0.95*0.95
=0.857375

P(1200 on time, here we make sure 1000 is on time(why ? because we want both to be on time)) = 0.95*0.05*0.6
=0.0285

P(A n B)= P(1000 on time AND 1200 on time) = 0.857375 + 0.0285
=0.885875

P(A|B) = P(A n B) / P (B)
P(A | B) = 0.885875 / 0.926375

P(A|B)= 0.956281...

I think this is the right method and answer, check the mark scheme though.

[pratstercs]

(Original post by bijesh12)
P(A|B)= 0.956281...

I think this is the right method and answer, check the mark scheme though.

Yeah, that is the right answer, but one bit I don't get:

P(1000 is on time) = 0.95*0.95*0.95
=0.857375

Why is this not just 0.95 (probability 1000 is on time, regardless of the next probabilities)?
Thanks!

[bijesh12]

(Original post by pratstercs)
Yeah, that is the right answer, but one bit I don't get:


Why is this not just 0.95 (probability 1000 is on time, regardless of the next probabilities)?
Thanks!

Remember both trains have to be on time, so 1000 has to be on time, in order for 1100 to be on time, in order to for 1200 to be on time.

We explore two possibilities for 1200 to be on time, 1) where all trains are on time and 2) where 1000 is on time, but 1100 is not on time.

[pratstercs]

(Original post by bijesh12)
Remember both trains have to be on time, so 1000 has to be on time, in order for 1100 to be on time, in order to for 1200 to be on time.

We explore two possibilities for 1200 to be on time, 1) where all trains are on time and 2) where 1000 is on time, but 1100 is not on time.

OK, I think I get it now. Thanks!

[pratstercs]

Right, I still don't entirely get it.
I'm doing June 09 S1 (PDF) Question 7iv:
Given that the flight is delayed, find the probability that the delay is down to just one of the 3 problems.
The prob that it is delayed is (1-0.612) = 0.388.
The prob it has only one problem is .
I do = but this gives an answer of 2.18 which is clearly not right. Where have I gone wrong?