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Ex 14.2
Ex 14.2, 2
Ex 14.2, 3 (i) Important
Ex 14.2, 3 (ii)
Ex 14.2, 3 (iii)
Ex 14.2, 3 (iv)
Ex 14.2, 3 (v)
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Ex 14.2, 11 Important
Ex 14.2, 12 (i)
Ex 14.2, 12 (ii) Important
Ex 14.2, 13 You are here
Ex 14.2, 14 Important
Ex 14.2, 15 Important
Ex 14.2, 16 Important
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Ex 14.2, 20 Important
Ex 14.2, 21 Important
Ex 14.2
Last updated at May 29, 2023 by Teachoo
Ex 14.2, 13 Fill in the blanks in following table: We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values P(A ∪ B) = 13 + 15 – 115 P(A ∪ B) = 5 + 3 − 115 P(A ∪ B) = 715 Hence P(A ∪ B) = 𝟕𝟏𝟓 Ex 14.2, 13 Fill in the blanks in following table: We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 0.6 = 0.35 + P(B) – 0.25 0.6 = 0.35 – 0.25 + P(B) 0.6 = 0.10 + P(B) 0.6 – 0.10 = P(B) 0.5 = P(B) P(B) = 0.5 Hence P(B) = 0.5 Ex 14.2, 13 Fill in the blanks in following table: We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 0.7 = 0.5 + 0.35 – P(A ∩ B) 0.7 = 0.85 – P(A ∩ B) P(A ∩ B) = 0.85 – 0.7 P(A ∩ B) = 0.15 Hence, P(A ∩ B) = 0.15