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Ex 16.3
Ex 16.3, 2
Ex 16.3, 3 (i) Important
Ex 16.3, 3 (ii)
Ex 16.3, 3 (iii)
Ex 16.3, 3 (iv)
Ex 16.3, 3 (v)
Ex 16.3, 4 Important
Ex 16.3 ,5 Important
Ex 16.3, 6
Ex 16.3, 7 Important
Ex 16.3, 8 Important
Ex 16.3, 9
Ex 16.3, 10
Ex 16.3, 11 Important
Ex 16.3, 12 (i)
Ex 16.3, 12 (ii) Important
Ex 16.3, 13 You are here
Ex 16.3, 14 Important
Ex 16.3, 15 Important
Ex 16.3, 16 Important
Ex 16.3, 17
Ex 16.3, 18
Ex 16.3, 19
Ex 16.3, 20 Important
Ex 16.3, 21 Important
Last updated at May 29, 2018 by Teachoo
Ex 16.3, 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 16.3, 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 16.3, 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