Oh, Probablility... Watch

nk9230
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Given P(A)=.6, P(A \land B')=.4, and P(A \lor B)=.85, Find the value of P(B).

I started by filling what was given into some formulae...

.85= .6 + P(B) - P(A \lor B)
.4= .6+ P(B') - P(A \lor B')

Am I on the right track?
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Kevin_B
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B or P(B)?
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nk9230
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(Original post by Kevin_B)
B or P(B)?
yeah, P(B), my mistake, I just edited.
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Kevin_B
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P( A and B' ) is just the intersection of A without B, draw a venn diagram and you can find the intersection, then use P( A or B ) = P(A) + P(B) - P( A and B ). Is this edexcel?
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ghostwalker
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(Original post by nk9230)
Given P(A)=.6, P(A \land B')=.4, and P(A \lor B)=.85, Find the value of P(B).

I started by filling what was given into some formulae...

.85= .6 + P(B) - P(A \lor B)
.4= .6+ P(B') - P(A \lor B')

Am I on the right track?

First equation should be:

.85= .6 + P(B) - P(A \land B)

For your second equation I would look to express the probability of A in terms of the two subsets A\land B\text{ and }A\land B'
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nk9230
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(Original post by Kevin_B)
P( A and B' ) is just the intersection of A without B, draw a venn diagram and you can find the intersection, then use P( A or B ) = P(A) + P(B) - P( A and B ). Is this edexcel?
No, IB Maths SL.
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nk9230
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(Original post by ghostwalker)
First equation should be:

.85= .6 + P(B) - P(A \land B)

For your second equation I would look to express the probability of A in terms of the two subsets A\land B\text{ and }A\land B'</b>
Yeah, I typed the equation in wrong :p: , and How would I do that?
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