Given that P(A) = 0.3 and P(B) = 0.9, and P(A and B) = 0.18, determine P(B|A). 0.30 0.60 0.15 0.20

Is P(A and B) \[P(A \cup B)?\]

yes

ok P(A) = 0.3 and P(B) = 0.9, and P(A and B) = 0.18 P(B\A)=P(B)-P(A intersection B) P(AUB)=P(A)+P(B)-P(A intersection B)--->P(A intersection B)=0.06 P(B\A)=0.9-0.06=0.84

Get it?

I think P(A and B) is supposed to be P(A n B) instead...since that answer isn't one of the choices...

I look at it from a Venn diagram view... P(B|A) is P(B given A)...so we know A has occurred. Using the shown diagram, we see that B, given A is P(A n B)/P(A) since A is the new sample space. So, \[P(B|A)=.18/.3 = .6\]

Ok then..sorry...Idk these words in english.... So if P(A and B) = 0.18 is the intersection we get ---> And P(B\A)=P(B)-P(A intersection B)=0.9-018=0.72

Still not one of the choices she gave.

K..these are the formulas I know..Idk....let me see again :)

The conditional probability formula is \[P(B|A)=P(B n A)/P(A)\] It asks for, what is the porbability that B has occurred, *given* the fact that A has occurred. So we're looking at what portion of B overlaps A.

Idk...sorry

No problem. I'm 99.99% sure the answer is .6

I've always thought that these formulas are same as when we're dealing with sets ...idk though

P(A) = 0.3; P(B) = 0.9, P(A and B) = 0.18 determine P(B|A). P(AnB) = P(A)*P(B) when they are mutally exclusive events P(B|A) = P(AnB)/P(A) when they are not mutually exclusive

Right, becuase P(B|A) = 0 if B and A are mutually exclusive

since P(AnB) = .27 if they were exclusive; and it says they equal .18 we know that they aint :) .18 18 --- = -- = 6/10 = .6 .3 30

Wait...P(AnB) = 0 when they are mutually exclusive, right? SInce there would be no overlap?

no; P(AnB) when they are mutually exclusive means that A doesnt affect B

Ok...you're right. THe formulas are still there, but the definitions went out the window years ago.

a.b --- = b when a and b are seperate entities a

I was thinking disjoint...

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