# Probabilty of AnB

#### brendan

Hi Guys,
I have been given the following Question.
Given that P(A) = 0.4 and P(AorB) = 0.9 Find P(B) if A and B are independent.

I know That if the're independent

P(A&B) = P(A)P(B)

P(AorB) = P(A) + P(B) - P(A&B)

And as the're independent

P(A|B) = P(A)

and

P(A|B) = P(A&B)/P(B)

As P(A&B) = P(A) + P(B) - P(A&B)

I know P(A) = 0.4, P(A|B) = 0.4 P(AorB)= 0.9 and

P(AorB) <= P(A) + P(B)
0.9 <= 0.4 + P(B)
Which implies that P(B) <= 0.5

I'm Trying to find P(B) however I'm having a lot of difficulty finding P(A&B).

Could someone please point me in the right direction of finding P(A&B)?

Kindest regards
Brendan

#### HallsofIvy

Hi Guys,
I have been given the following Question.
Given that P(A) = 0.4 and P(AorB) = 0.9 Find P(B) if A and B are independent.

I know That if the're independent

P(A&B) = P(A)P(B)

P(AorB) = P(A) + P(B) - P(A&B)

And as the're independent

P(A|B) = P(A)

and

P(A|B) = P(A&B)/P(B)

As P(A&B) = P(A) + P(B) - P(A&B)
Typo: that last "P(A&B)" should be P(AorB)

I know P(A) = 0.4, P(A|B) = 0.4 P(AorB)= 0.9 and

P(AorB) <= P(A) + P(B)
0.9 <= 0.4 + P(B)
Which implies that P(B) <= 0.5

I'm Trying to find P(B) however I'm having a lot of difficulty finding P(A&B).

Could someone please point me in the right direction of finding P(A&B)?

Kindest regards
Brendan
You have already said that P(A&B)= P(A)P(B) because they are independent.
P(AorB)= P(A)+ P(B)- P(A&B)= P(A)+ P(B)- P(A)P(B).

You are told that P(A)= 0.4 and P(AorB)= 0.9. That equation becomes
0.9= 0.4+ P(B)- 0.4P(B). Solve that equation.

#### brendan

Thanks alot for your help I really appreciate it
0.9= 0.4+ P(B)- 0.4P(B). Solve that equation.

p(B) = .833 and p(A&B) = .33

I will be remembering that one!

regards
Brendan

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