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) \(P(\bar A ∩B)\) (iv) \(P(A ∩ \bar B)\)

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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