2630101
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Okay so if you are given the probability of A and also P(B), then why does the example work it out from the equation P(A|B)??
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2630101
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(Original post by 2630101)
Okay so if you are given the probability of A and also P(B), then why does the example work it out from the equation P(A|B)??
Trying to upload the example now
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Didn't really help but thanks anyways
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Miss.Modesty
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Because thats just how the formula works.

P(A|B)= P(AnB)/P(B)
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PuffyPrincess
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(Original post by 2630101)
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It's just like plugging some numbers into an equation any other way. You just need to rearrange it! You know that the probability of A given B= Probability (A and B) over Probability B... So just plug it in. I mean, you know it's that equation since the first line tells you what the probability of A given B is. Not sure what you're struggling with! The equation makes sense logically if you think about it, you're looking for the prob. of A happening when B has already happened, so you look at the probability they both happen, but since B has already happened you can take that probability out by dividing.
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headyb
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It's really just division.

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2630101
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(Original post by headyb)
It's really just division.

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I understand how to do all the rearranging the formula, but why cant i just do 0.5x0.9??
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PuffyPrincess
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(Original post by 2630101)
I understand how to do all the rearranging the formula, but why cant i just do 0.5x0.9??
Because they're not independant. So P(A and B) is not the same as P(A) x P(B).
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headyb
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(Original post by 2630101)
I understand how to do all the rearranging the formula, but why cant i just do 0.5x0.9??
Aha. That would be the probability of event A happening, then event B happening, or visa versa.

It depends on the situation if you should use this. If the probability of B changes given A has changed, you can't just multiple them together to find both probabilities (as the values are no longer the same!)

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For instance, if you were to toss a coil and roll a die,
P(Heads) = 0.5
P(6 on die) = 1/6

You could work out the probability of getting heads then a 6. You would multiply them together.
P(Heads AND 6 on die) = 0.5 x 1/6 = 1/12.

But it wouldn't make any sense trying to find the probability of getting a 6 given you have already got a heads. This is because the probabilities aren't affected by one another (they're independent).
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NotNotBatman
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(Original post by 2630101)
Okay so if you are given the probability of A and also P(B), then why does the example work it out from the equation P(A|B)??
P(A|B) is the probability of A given B, so you can think of the entire probability set as B and work out how much of A is contained within that set. So  \frac{P(A \cap B)}{P(B)}
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TheNumbere
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You can't just multiply because the probabilities depend on each other. Ie. They are related. This is why simply multiplying won't work.
Multiplying out would work if these two events were independent. You really should know this.
And also is this aGCSE question or As Question? I ask out of pure curiosity!

we are told that P(A|B)= 0.72
We are told that P(B)=0.5

simply substitute values in the formula: P(A|B)=P(A And B)/ P(B)
we get: 0.72= P(A And B)/0.5 ====> 0.36
Then that answers Part A of the question

Part B is virtually the same. You just change numbers.
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(Original post by TheNumbere)
You can't just multiply because the probabilities depend on each other. Ie. They are related. This is why simply multiplying won't work.
Multiplying out would work if these two events were independent. You really should know this.
And also is this aGCSE question or As Question? I ask out of pure curiosity!

we are told that P(A|B)= 0.72
We are told that P(B)=0.5

simply substitute values in the formula: P(A|B)=P(A And B)/ P(B)
we get: 0.72= P(A And B)/0.5 ====> 0.36
Then that answers Part A of the question

Part B is virtually the same. You just change numbers.
Further Maths GCSE, and thanks
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