Given P(A∪B)=0.7, P(A|B')=0.6, find P(B)

Answer 1

Hi, Muhong!

Well... It's all about using some probabilities rules.

Please, check it out:

i. P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.7; it means that P(A ∩ B) = P(A) + P(B) - 0.7

ii. P(A | B') = P(A ∩ B') / P(B') = [P(A) - P(A ∩ B)] / [1 - P(B)] = 0.6

Now, let's plug (i) in (ii):

[P(A) - P(A ∩ B)] / [1 - P(B)] = 0.6

[ P(A) - [P(A) + P(B) - 0.7] ] / [1 - P(B)] = 0.6

[ P(A) - P(A) - P(B) + 0.7 ] / [1 - P(B)] = 0.6

Cancelling P(A) we got: [ - P(B) + 0.7 ] / [1 - P(B)] = 0.6

that means [ - P(B) + 0.7 ] = 0.6 * [1 - P(B)].

distribute the right side to get

- P(B) + 0.7 = 0.6 - 0.6 * P(B)

now, isolating P(B) we got:

- P(B) + 0.6 * P(B) = 0.6 - 0.7

-0.4 * P(B) = -0.1

Finally P(B) = -0.1 / -0.4 = 0.25 = 25%

Let me know if you need more help, ok?!

See you!!

Answer 2

It turns out to be 0.1

Drawing the Venn Diagram helps