Q. 184.2( 13 Votes )

Two events A and (A) A and B are mutually exclusive

(B) P(A′B′) = [1 – P(A)] [1 – P(B)]

(C) P(A) = P(B)

(D) P(A) + P(B) = 1

Answer :

Given: Two events A and B will be independent.

As A and B are independent events.


P (A B) = P(A) . P(B)


We solve it using options.


Let P(A) = a, P(B) = b


As, A and B are mutually exclusive


P (A B) = ϕ


Now, P(A).P(B) = a.b ≠ P(A B)


P (A B) P(A) . P(B)


Hence, it shows A and B are not Independent events.


(B) P(A′B′) = [1 – P(A)] [1 – P(B)]


P(A′ B) = 1 P(A) P(B) + P(A)P(B)


1 - P (A B) =1 – P(A) – P(B) + P(A)P(B)


= - [P(A) + P(B) - P (A B)] = P(A) P(B) + P(A)P(B)


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


P (A B) = P(A) . P(B)


Hence, it shows A and B are Independent events.


(C) P(A) = P(B)


As, P(A) = P(B)


Let we take the example of a coin


P(A) = probability of getting head = 1/2


P(B) = probability of getting tail = 1/2


A B = ϕ


P(A B) = probability of getting head and tail both = 0


Now, P(A).P(B) = 1/2 . 1/2 = 1/4 ≠ P(A B)


P (A B) P(A) . P(B)


Hence, it shows A and B are not Independent events.


(D) P(A) + P(B) = 1


Let we take the example of a coin


P(A) = probability of getting head = 1/2


P(B) = probability of getting tail = 1/2


Now, P(A) + P(B) = 1/2 + 1/2 = 1


But it doesnot inferred that A and B are independent.


Hence the correct option is B.

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