Ex 16.3, 13 - Chapter 16 Class 11 Probability (Term 2)
Last updated at May 29, 2018 by
Last updated at May 29, 2018 by
Transcript
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
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Ex 16.3, 13 You are here
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