Answer to Question #131455 in Statistics and Probability for Ajay

Question

If p1=p(A),p2=p(B),p3=P(A ∩ B), (p1,p2,p3>0) express the following in terms of p1,p2,p3

1)P(A ∪ B)'
2)P(A'∪ B')
3)P(A' ∩ B)
4)P(A' ∪ B)
5)P(A' ∩ B')
6)P(A ∩ B')
7)P(A | B)
8)P(B | A')
9)P[A' ∩ (A ∪ B)]

Answer 1

1)

$P(A ∪ B)' = 1-P(A \cup B) = 1 - [P(A)+P(B)-P(A \cap B)]$

$\therefore P(A ∪ B)' = 1-(p1+p2-p3) = 1-p1-p2+p3$

2)

$P(A'∪ B')=1-P(A\cap B) = 1-p3$

3)

$P(A' \cap B)=P(B)-P(A\cap B)=p2-p3$

4)

$P(A' ∪ B)=1-P(A\cap B')=1-[P(A)-P(A\cap B)]$

$P(A' ∪ B)=1-(p1-p3)=1-p1+p3$

5)

$P(A' \cap B') = 1-P(A \cup B) = 1 - [P(A)+P(B)-P(A \cap B)]$

$\therefore P(A' \cap B') = 1-(p1+p2-p3) = 1-p1-p2+p3$

6)

$P(A \cap B')=P(A)-P(A\cap B)=p1-p3$

7)

$P(A | B)= \frac {P(A\cap B)}{P(B)}=\frac {p3}{p2}$

8)

$P(B | A')=\frac {P(B\cap A')}{P(A')}=\frac {P(B)-P(A\cap B)}{1-P(A)}= \frac {p2-p3}{1-p1}$

9)

$P[A' ∩ (A ∪ B)]=P(B\cap A')=P(B)-P(A\cap B)$

$P[A' ∩ (A ∪ B)]=p2-p3$