A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩

1.	A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) =0.35. Find (i) P(A ∪ B)(ii) \(P(\bar A ∩ \bar B)\)(iii) \(P(A ∩ \bar B)\) (iv) \(P(B∩ \bar A)\)
Answer» Given A and B are two events And, P(A) = 0.54 P(B) = 0.69 P(A ∩ B) = 0.35 By definition of P(A or B) under axiomatic approach we know that: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) We have to find (i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.54 + 0.69 – 0.35 = 0.88 (ii) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law} ⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.88 = 0.12 (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) ∴ P(A ∩ B’) = P(A) - P(A ∩ B) = 0.54 – 0.35 = 0.19 (iv) 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) ∴ P(A’ ∩ B) = P(B) – P(A ∩ B) = 0.69 – 0.35 = 0.34

A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) =0.35. Find (i) P(A ∪ B)(ii) \(P(\bar A ∩ \bar B)\)(iii) \(P(A ∩ \bar B)\) (iv) \(P(B∩ \bar A)\)

Answer»

Given A and B are two events

And, P(A) = 0.54 P(B) = 0.69 P(A ∩ B) = 0.35

By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We have to find

(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.54 + 0.69 – 0.35 = 0.88

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

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.88 = 0.12

(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)

∴ P(A ∩ B’) = P(A) - P(A ∩ B) = 0.54 – 0.35 = 0.19

(iv) P(A’ ∩ B) = This indicates only the part which is common with B and not A ⇒ This indicates only B.