Stuck on probability question: (A and B are independent) P(B ∩ [not A] | A ∪ B)

leander

New member
Joined
Jun 25, 2018
Messages
1
Stuck on probability question: (A and B are independent) P(B ∩ [not A] | A ∪ B)

I'm having a bit of trouble understanding on how to get from this
(A and B are independent)


P(B ∩ [not A] | A ∪ B)


to this


P(B ∩ [not A]) / P(A ∪ B)


My progress:


P[(B ∩ [not A]) ∩ (A ∪ B)] / P(A ∪ B) =


= P[(B ∩ [not A] ∩ A) ∪ (B ∩ [not A] ∩ B)] / P(A ∪ B) =


= ? (now i'm stuck)
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,349
I'm having a bit of trouble understanding on how to get from this
(A and B are independent)

P(B ∩ [not A] | A ∪ B)

to this

P(B ∩ [not A]) / P(A ∪ B)

My progress:

P[(B ∩ [not A]) ∩ (A ∪ B)] / P(A ∪ B) =

= P[(B ∩ [not A] ∩ A) ∪ (B ∩ [not A] ∩ B)] / P(A ∪ B) =

= ? (now i'm stuck)
Your goal is to show that (B ∩ [not A]) ∩ (A ∪ B) = B ∩ [not A]. I would just write it like that, rather than keep writing P's.

You have taken a good first step by distributing. Now look closely at each part of the resulting union, and apply the commutative and associative properties:

B ∩ [not A] ∩ A = B ∩ ([not A] ∩ A): Do you see what [not A] ∩ A is?

Then put that result into (...) ∩ (A ∪ B) and see what you get. (Another distribution may help.)

It may help if you keep a list of properties in front of you as you do this, so you can scan it at each step to see if anything is relevant. What properties have you learned?

By the way, there are many notations for "not"; one that works well in typing is ~A for not A, and another is A'. That may make things a little easier. I suppose you probably have been taught the overbar.
 
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