X


Payment

Question

let A, B and C be events in a sample space and P be a probability. Prove the following:

(a) If A and B and A and C are independent and moreover [B\cap C=\varnothing] then A and [B\cup C] are independent as well.

(b) Assume not that A, B and C are indepenedent. Then A and [B\cup C] are independent

Solution

(x) If x xxx x and x and x xxx xxxxxxxxxxx

=> x(x x) = x(x) P(B)

x(x x) = x(x) P(C)

xxxx P(B x) = 0

=> P(B x x) = P(B) + P(C)

x(x (x U x)) = x(x) + x(x U x) - x(x x x U x)

= x(x) + x(x) + x(x) - x(x x x U x)

= x(x) + x(x) + x(x) - (x(x) + x(x) + x(x) - x(x x) - x(x x) - x(x x) + x(x x x))

= x(x x) + x(x x) + x(x x) - x(x x x)

= x(x)x(x) + x(x)x(x) + x - x = x(x) (P(B) + P(C))

= x(x) x(x U x)

Therefore x xxx x U x are xxxxxxxxxxx. (x) (xxxx does xxxxxx not xxxx? xxx xxx is xxxxxxxy mistakenly xxxxx) x(x (B x C)) = x(x) + P(B x C) - x(x x B x C)

= x(x) + (P(B) + P(C) - x(x C)) - P(A x x x C)

= P(A) + (x(x) + P(C) - P(B x)) - (P(A) + P(B) + x(x) - P(A B) - x(x C) - P(B x) + P(A B x)) = P(A) + P(B) + x(x) - P(B C) - (x(x) + P(B) + P(C) - x(x B) - P(A x) - P(B C) + x(x B C))

= x(x B) + P(A x) - P(A B x) = P(A) x(x) + x(x)x(x) - x(x) P(B) x(x)

= x(x) (x(x) + P(C) - P(B) x(x)) = x(x) P(B x C)

xxxxxxxxx x xxx B xxx independent.

Payment



Payment