Question

For events A and B, the following probabilities are known: Pr[A!B]=.8 andPr[A]=.5
What is the value of Pr[B]if A and B are disjoint (this means the events are partitions)?

Answer 1

Any help would be great. An explanation would help

Answer 2

is that Pr(A|B)=0.8?

Answer 3

Pr[AUB] = .8 and Pr[A] = .5

Answer 4

sorry

Answer 5

trying to find Pr[b]

Answer 6

and yes I believe

Answer 7

Oh ok makes more sense now.
\(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

Since the events are disjoint, \(P(A\cap B)=0\)

So \(P(B)=P(A\cup B)-P(A)\)

Answer 8

ok so that is what I thought. it is equal to zero because it is disjoint. so really no resolution. correct?

Answer 9

or that is the equation we have to work with? But yes that is what I am asking

Answer 10

Um that is really just the formula for unions of events. And the intersection is 0 because A and B are disjoint. So you just solve for your missing variable, P(B).

Answer 11

yes exactly which I can not figure out the method of finding the answer. having a hard time with this

Answer 12

So P(B)=P(A∪B)−P(A) is our starting point so really are you saying .8-.5 would equal Pr(B)

Answer 13

yes

Answer 14

is that not the right answer?

Answer 15

ahh you sure

Answer 16

not sure. trying to find it out. that is what i had to work with

Answer 17

Ok. well that should be the answer then :P

Answer 18

wow so it really is 0

Answer 19

thank you

Answer 20

Yes. By definition of disjoint, \(P(A \cap B)=0\)

Answer 21

thanks again

Answer 22

make sense now

Answer 23

:)