If A and B be mutually exclusive events associated with a random exper

1.	If A and B be mutually exclusive events associated with a random experiment such that P (A) = 0.4 and P (B) = 0.5, then find:(i) P(A ∪ B)(ii) P(bar A ∩ bar B)(iii) P(bar A ∩ bar B)(iv) P(A ∩ bar B)
Answer» Given: A and B are two mutually exclusive events. P (A) = 0.4 and P (B) = 0.5 By definition of mutually exclusive events we know that: P (A ∪ B) = P (A) + P (B) Now, we have to find (i) P (A ∪ B) = P (A) + P (B) = 0.5 + 0.4 = 0.9 (ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law} P (A′ ∩ B′) = 1 – P (A ∪ B) = 1 – 0.9 = 0.1 (iii) P (A′ ∩ B) [This indicates only the part which is common with B and not A. Hence this indicates only B] P (only B) = P (B) – P (A ∩ B) As A and B are mutually exclusive so they don’t have any common parts. P (A ∩ B) = 0 ∴ P (A′ ∩ B) = P (B) = 0.5 (iv) P (A ∩ B′) [This indicates only the part which is common with A and not B. Hence this indicates only A] P (only A) = P (A) – P (A ∩ B) As A and B are mutually exclusive so they don’t have any common parts. P (A ∩ B) = 0 ∴ P (A ∩ B′) = P (A) = 0.4

If A and B be mutually exclusive events associated with a random experiment such that P (A) = 0.4 and P (B) = 0.5, then find:(i) P(A ∪ B)(ii) P(bar A ∩ bar B)(iii) P(bar A ∩ bar B)(iv) P(A ∩ bar B)

Answer»

Given: A and B are two mutually exclusive events.

P (A) = 0.4 and P (B) = 0.5

By definition of mutually exclusive events we know that:

P (A ∪ B) = P (A) + P (B)

Now, we have to find

(i) P (A ∪ B) = P (A) + P (B) = 0.5 + 0.4 = 0.9

(ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.9

= 0.1

(iii) P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence this indicates only B]
P (only B) = P (B) – P (A ∩ B)

As A and B are mutually exclusive so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A′ ∩ B) = P (B) = 0.5

(iv) P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence this indicates only A]

P (only A) = P (A) – P (A ∩ B)

As A and B are mutually exclusive so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A ∩ B′) = P (A) = 0.4