1.

If A and B are mutually exclusive events such that P(A) = 0.35 and P(B) = 0.45, find (i) P (A∪B) (ii) P (A∩B)(iii) \(P (A∩\bar{B})\)(iv) \(P (\bar{A}∩\bar{B})\)

Answer»

Given A and B are two mutually exclusive events 

And, P(A) = 0.35 P(B) = 0.45 

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) = 0.35 + 0.45 = 0.8 

ii) P(A ∩ B) = 0 {∵ nothing is common between A and B} 

iii) 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) = 0.35 

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

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.8 = 0.2


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