| P(AUB) = P(A) + P(B) - P(ANB) | P(B|A) = P(ANB) / P(A)
P(A|B) = P(ANB) / P(B) | P(exactly 1 of A and B) =
P(A) + P(B) - 2P(AnB) |
"The probability of the Union of A and B
is equal to Prob A plus Prob B
- and just in case they are not mutually exclusive
we must knock off any duplicates
so minus the Prob of A and B" | The probability of B occuring
given that A has already occured
or
The probability of A occuring
given that B has already occured
| "The probability of exactly ONE of A and B occuring can be calculated using this formula"
It is actually P(AUB) - P(AnB) but ..
since P(AUB) = P(A) + P(B) - P(AnB)
it becomes
P(AUB) = P(A) + P(B) - P(AnB) - P(AnB) |
This is the main one
remember this and you are half way there | This one comes up in exam papers all the time
Just remember it and pick up the points | This one confuses me.
I have no idea what the question means.
Ask me about red balls and green balls, or rolling dice or men with beards and I can usually understand it.
But this one leaves me clueless.
|
Don't confuse "mutually exclusive" with "independent".
"mutually exclusive" means that
none of group A are also in group B
"independent" means that
the number you would normally expect
are in both groups.
For example, if 50% of a group are women
and 50% of the same group are married
then you would expect 25% to be married women, if they were independent. | If B depend on A, then A depends on B and :
The probability of A occuring given that B has already occured can be calculated using :
(note P(A1) is the probability of A not happening and P(B|A1) is the probability of B if A doesn't happen)
P(A|B) = P(ANB) / P(B) = P(B|A) * P(A) / (P(B|A) * P(A) + P(B|A1) * P(A1)) | So I don't understand the question,
but I do know how to work out the answer.
Am I dumb or am I genius?
You do the Math
I think I may have sorted this out.
Have a look at my
Understanding Probabilities
(near the bottom)l |