Ex 16.3, 15 - If P(E) =  1/4, P(F) = 1/2, P(E and F) = 1/8

Ex 16.3, 15 - Chapter 16 Class 11 Probability - Part 2
Ex 16.3, 15 - Chapter 16 Class 11 Probability - Part 3

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Transcript

Ex 14.2, 15 If E & F are events such that P(E) = 1/4 , P(F) = 1/2 and P(E and F) = 1/8, find: (i) P(E or F) P(E and F) = P(E ∩ F) = 1/8 We need to find P(E or F) = P(E ∪ F) We know that P(E ∪ F) = P(E) + P(F) – P(E ∩ F) Putting values P(E ∪ F) = 1/4 + 1/2 – 1/8 = (2 + 4 − 1)/8 = (6 − 1)/8 = 𝟓/𝟖 Ex 14.2, 15 If E & F are events such that P(E) = 1/4 , P(F) = 1/2 and P(E and F) = 1/8, find: (ii) P(not E and not F). P (not E and not F) = P(E’ ∩ F’) = P (E ∪ F)’ = 1 – P (E ∪ F) = 1 – 5/8 = (8 − 5)/8 = 𝟑/𝟖 Demorgan’s law █("If (A’" ∩"B’) = (A " ∪" B)’ " @"or (A’ " ∪" B’) = (A " ∩" B)’" )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.