Ex 14.2, 17 - Chapter 14 Class 11 Probability
Last updated at April 16, 2024 by Teachoo
Ex 14.2
Ex 14.2, 1
Ex 14.2, 2
Ex 14.2, 3 (i) Important
Ex 14.2, 3 (ii)
Ex 14.2, 3 (iii)
Ex 14.2, 3 (iv)
Ex 14.2, 3 (v)
Ex 14.2, 4 Important
Ex 14.2 ,5 Important
Ex 14.2, 6
Ex 14.2, 7 Important
Ex 14.2, 8 Important
Ex 14.2, 9
Ex 14.2, 10
Ex 14.2, 11 Important
Ex 14.2, 12 (i)
Ex 14.2, 12 (ii) Important
Ex 14.2, 13
Ex 14.2, 14 Important
Ex 14.2, 15 Important
Ex 14.2, 16 Important
Ex 14.2, 17 You are here
Ex 14.2, 18
Ex 14.2, 19
Ex 14.2, 20 Important
Ex 14.2, 21 Important
Chapter 14 Class 11 Probability
Serial order wise
- Ex 14.1
-
Ex 14.2
- Examples
- Miscellaneous
- Sample Space
Transcript
Ex 14.2, 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 14.2, 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 14.2, 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
