# 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 υ B) (ii) P ( A' ∩ B') (iii) P ( A ∩ B') (iv) P ( B ∩ A')

**Solution:**

It is given that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35.

**(i)** By P (A υ B) formula,

P (A υ B) = P (A) + P (B) - P (A ∩ B)

= 0.54 + 0.69 – 0.35

= 0.88

**(ii)** By De Morgan’s law,

(A' ∩ B') = (A υ B)'

Therefore,

P (A' ∩ B') = P (A υ B)'

= 1 - P (A υ B)

= 1 – 0.88 (from (i))

= 0.12

**(iii)** We know that

P (B ∩ A') = P (A) - P (A ∩ B)

= 0.54 - 0.35

= 0.19

**(iv)** We know that

n (B ∩ A') = n (B) - n (A ∩ B)

Therefore,

P (B ∩ A') = P (B) - P (A ∩ B)

= 0.69 - 0.35

= 0.34

NCERT Solutions Class 11 Maths Chapter 16 Exercise ME Question 7

## 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 υ B) (ii) P ( A' ∩ B') (iii) P ( A ∩ B') (iv) P ( B ∩ A').

**Summary:**

A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35. Then (i) P(A υ B) = 0.88 (ii) P(A' ∩ B') = 0.12 (iii) P(B ∩ A') = 0.19 (iv) P(B ∩ A') = 0.34

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