A and B are discrete events. You are given values for P(A) and P(B). What value of P(NOT A NOT B) will make A and B independent?

Could somebody help me to solve this problem? Thank you.

P(not A not B)=1-P(AB)-P(notA ,B)-P(A,not B)
If A and B are to be independent
P(not A not B)=1-P(A)P(B)-P(not A)P(B)-P(A)P(not B)
I think this should be the answer

Perrybhadra is right, but if it helps you understand better, here is another way:

P(not A and not B)
=1- P (A union B)
= 1 - [ P(A) + P(B) - P(A and B)
= 1- P(A) -A(B) + P(A)P(B) (if A and B are independent)

*you should get the same answer both ways.

THANKS GUYS.

Do you guys know how to solve this one?


A,B,C and D are discrete events. P(AB+CD) = P(AB) + P(CD); P(BD) = 0.6; P(NOTA BD) = 0.4.
Using the Veitch diagram, find P(AB NOTC D).

Thank you.

I solved the second one already. I got 0.2 for answer.


P(B|NOTA) = ???
Given P(A), P(B) and P(AB).
A and B are not independent.

Seems like you need to read the chapter few more times, Pra. :) These questions ask for basic definitions in probability. If you are not clear with it, may be, it would be helpful for you to go over few more times.

P(B/notA)
= P(B and notA)/P(notA)
=[P(B) - P(A and B)]/[1-P(A)]

I took probability class 4 yrs ago and now it is the base for constraint theory class. Professor gave us these problems without explaining anything and it’s not covered on the book either. I don’t have my old books anymore. Thanks for your help.