Given P(B | A)=0.75, P(A ∩ B)=0.15, P(B’)=0.7, find P(A ∪ B)

ince we have given, P(A) = 0.75 and P(A|B) =0.8 .

So, P(A∩B) = P(B|A)× P(A) = 0.8×0.75 =0.6 .

We have given P(B|A') = 0.6

P(A’∩B) = P(B|A')×P(A') = 0.6 × 0.25 = 0.15.

P(B) = P(A∩B) + P(A'∩B) = 0.6 + 0.15 = 0.75.

=> P(A|B) =  P(A∩B)P(B)=0.60.75  = 0.8 .

Step-by-step explanation:

By definition of conditional probability,

P(B | A) = P(A ∩ B) / P(A)

==>   P(A) = P(A ∩ B) / P(B | A) = 0.15/0.75 = 0.2

By definition of complement,

P(B') = 1 - P(B)

==>   P(B) = 1 - P(B') = 1 - 0.7 = 0.3

Now by the inclusion/exlcusion principle, we have

P(A U B) = P(A) + P(B) - P(A ∩ B)

==>   P(A U B) = 0.2 + 0.3 - 0.15 = 0.35

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

its kind of blurry. can you put a clearer pic?

step-by-step explanation: