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. P (A) = 0.54, P (B) = 0.69 and 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) Now 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. Hence 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. Hence 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.

P (A) = 0.54, P (B) = 0.69 and 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)

Now 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.

Hence 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.

Hence 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