Q. 1 A5.0( 1 Vote )
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) 
(iii) 
(iv) 
Answer :
Given A and B are two mutually exclusive events
And, P(A) = 0.4 P(B) = 0.5
By definition of mutually exclusive events we know that:
P(A ∪ B) = P(A) + P(B)
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 ⇒ 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 ⇒ 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
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