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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