Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg
Ex 16.3, 17 - Chapter 16 Class 11 Probability
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 16.3, 17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), P(A) = 0.42 P(not A) = 1 P(A) = 1 0.42 = 0.58 Ex 16.3, 17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (ii) P (not B) P(B) = 0.48 P(not B) = 1 P(B) = 1 0.48 = 0.52 Ex 16.3, 17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (iii) P(A or B). P(A or B) = P(A B) We know that P(A B) = P(A) + P(B) P(A B) Putting values P(A B) = 0.42 + 0.48 0.16 = 0.74 Hence, P(A or B) = 0.74
Ex 16.3
Ex 16.3, 1
Ex 16.3, 2
Ex 16.3, 3 Important
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 Important
Ex 16.3, 13
Ex 16.3, 14 Important
Ex 16.3, 15 Important
Ex 16.3, 16 Important
Ex 16.3, 17 You are here
Ex 16.3, 18
Ex 16.3, 19
Ex 16.3, 20 Important
Ex 16.3, 21 Important
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.