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Ex 14.2, 13 - Chapter 14 Class 11 Probability
Last updated at Dec. 16, 2024 by Teachoo
Chapter 14 Class 11 Probability
Concept wise
Transcript
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
