Is P(A-B) = P(A) -P(B)Here p is the probability function

[crotical]

Is
P(A-B) = P(A) -P(B)

Here p is the probability function
Please help

 

[HallsofIvy]

If B is a subset of A then, yes. However, if B is not a subset of A, in which case "A- BZ means "all members of A that are not also in B", this is not necessarily true. For example, suppose we choose a number from 1 to 100, each number being equally likely. Let A be the set of all multiples of 3, Be the set of all multiples of 6. There are 100/3= 33 multiples of 3 so P(A)= 0.33. There are 100/6= 16 multiples of 6 so P(B)= 16/100= 0.16. Of course, all multiples of 6 are multiples of 3 so A- B is just all multiples of 3 that are NOT multiples of 6 and there are 33- 16= 17 such numbers. P(A- B)= 17/100= 0.17= 0.33- 0.16= P(A)- P(B).

But suppose that B, instead of being "multiples of 6" is "even numbers (multiples of 2)". We still have P(A)= 0.33 but now P(B)= 0.5. Of course it is impossible that "P(A- B)= P(A)- P(B)" because P(A)- P(B)= 0.33- 0.5= -0.17 and a probability cannot be negative. In fact, "A- B" would mean removing from A all those numbers that are even and multiples of 2 which is the same as "multiples of 6". For this A and B, we would still have P(A-B)= 0.17 which is, as I said, NOT "P(A)- P(B)".

 

[crotical]

Is this proof valid
Given P(A) and P(B) are independent , prove P(A) and P(B') independent too.
P(A∩B) = P(A)P(B)
P(A∩B) = P(A)P(S-B')
=P(A)(1-P(B'))
=P(A)-P(A)P(B')
P(A∩B)=P(A)-P(A)P(B')
P(A∩(S-B'))=P(A)-P(A)P(B')
P(A-A∩(B'))=P(A)-P(A)P(B')
P(A)-P(A∩B')=P(A)-P(A)P(B')
P(A∩B')=P(A)P(B')

 

[crotical]

I took S here as universal set

 

[Ray Vickson]

Is this proof valid
Given P(A) and P(B) are independent , prove P(A) and P(B') independent too.
P(A∩B) = P(A)P(B)
P(A∩B) = P(A)P(S-B')
=P(A)(1-P(B'))
=P(A)-P(A)P(B')
P(A∩B)=P(A)-P(A)P(B')
P(A∩(S-B'))=P(A)-P(A)P(B')
P(A-A∩(B'))=P(A)-P(A)P(B')
P(A)-P(A∩B')=P(A)-P(A)P(B')
P(A∩B')=P(A)P(B')

You can shorten it a bit: $A = A\cap[B \cup B'] = (A \cap B) \cup (A \cap B'),$ and these last two sets are mutually exclusive. Thus $P(A) = P(A \cap B) + P(A \cap B'), \text{ hence } P(A \cap B') = P(A) - P(A) \cdot P(B) = P(A) \cdot P(B').$

RGV