NCERT Class XI Mathematics - Probability - Solutions
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Question : 52
Total: 54
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(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′).
Solution:
Here P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
(i) We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.54 + 0.69 – 0.35 = 1.23 – 0.35 = 0.88
(ii) P(A′ ∩ B′) = P
(iii) P(A ∩ B′) = P(A) – P(A ∩ B) = 0.54 – 0.35 = 0.19
(iv) P(B ∩ A′) = P(B) – P(A ∩ B) = 0.69 – 0.35 = 0.34.
(i) We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.54 + 0.69 – 0.35 = 1.23 – 0.35 = 0.88
(ii) P(A′ ∩ B′) = P
(iii) P(A ∩ B′) = P(A) – P(A ∩ B) = 0.54 – 0.35 = 0.19
(iv) P(B ∩ A′) = P(B) – P(A ∩ B) = 0.69 – 0.35 = 0.34.
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