# Conditional Probability (confusing)

1. Sep 11, 2008

### sampahmel

Dear all,

P (A |B) + P (A c|B) = 1 [A c] denotes complement of set A and of course P (B)>0

Is the above statement true?

P (A |B) + P (A |B c) = 1

P (C ∪ D|B) = P (C |B) + P (D|B) − P (C ∩ D|B)

2. Sep 11, 2008

### NoMoreExams

What steps have you taken in proving/disproving these?

3. Sep 11, 2008

### rbeale98

first statement is true

P (A |B) + P (A |B c) = 1 is not true

if P(A) = x and A is independent of B, then P(A|B) = P(A|Bc) = x

i think the third statement is true

4. Sep 12, 2008

### Focus

TIP:
$$Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)} \quad Pr(B)>0$$
$$Pr(A \cup B)=Pr(A)+Pr(B)-Pr(A \cap B)$$