The symmetric difference between two events A and B P(A B) Prove that P(A B) = P(A) + P(B) - 2P(A B) Solution denoting symmetric difference as A~B We know that A~B = (A-B)U(B-A) Now A-B,B-A are disjoint events P(A~B) = P((A-B)U(B-A)) = P(A-B)+P(B-A) From Venn diagrams we can say that, P(A-B) = P(A) - P(A^B) (A^B denotes A intersection B) P(B-A) = P(B) - P(A^B) (A^B denotes A intersection B) Hence P(A~B) = P(A)+P(B)-2P(A^B).

1. The symmetric difference between two events A and B P(A B) Prove that P(A B) = P(A) +
P(B) - 2P(A B)
Solution
denoting symmetric difference as A~B
We know that A~B = (A-B)U(B-A)
Now A-B,B-A are disjoint events
P(A~B) = P((A-B)U(B-A)) = P(A-B)+P(B-A)
From Venn diagrams we can say that,
P(A-B) = P(A) - P(A^B) (A^B denotes A intersection B)
P(B-A) = P(B) - P(A^B) (A^B denotes A intersection B)
Hence P(A~B) = P(A)+P(B)-2P(A^B)