# Ex 13.4, 9 - Chapter 13 Class 12 Probability (Term 2)

Last updated at July 5, 2019 by Teachoo

Ex 13.4

Ex 13.4, 1

Ex 13.4, 2

Ex 13.4, 3 Important

Ex 13.4, 4 (i)

Ex 13.4, 4 (ii)

Ex 13.4, 4 (iii) Important

Ex 13.4, 5 (i) Important

Ex 13.4, 5 (ii)

Ex 13.4, 6 Important

Ex 13.4, 7 Important

Ex 13.4, 8

Ex 13.4, 9 You are here

Ex 13.4, 10 Deleted for CBSE Board 2022 Exams

Ex 13.4, 11 Important Deleted for CBSE Board 2022 Exams

Ex 13.4, 12 Important Deleted for CBSE Board 2022 Exams

Ex 13.4, 13 Important Deleted for CBSE Board 2022 Exams

Ex 13.4, 14 Deleted for CBSE Board 2022 Exams

Ex 13.4, 15 Deleted for CBSE Board 2022 Exams

Ex 13.4, 16 (MCQ) Deleted for CBSE Board 2022 Exams

Ex 13.4, 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Chapter 13 Class 12 Probability (Term 2)

Serial order wise

Last updated at July 5, 2019 by Teachoo

Ex 13.4, 9 The random variable X has a probability distribution P(X) of the following form, where k is some number : P(X) = ๐, ๐๐ ๐ฅ=0๏ทฎ2๐, ๐๐ ๐ฅ=1๏ทฎ3๐, ๐๐ ๐ฅ=2๏ทฎ0, ๐๐กโ๐๐๐ค๐๐ ๐๏ทฏ๏ทฏ (a) Determine the value of k. Making in tabular format Since X is a random variable , its Sum of Probabilities is equal to 1 0๏ทฎ4๏ทฎ๐(๐)๏ทฏ = 1 P(x = 0) + P(x = 1) + P(x = 2) = 1 k + 2k + 3k = 1 6k = 1 k = ๐๏ทฎ๐๏ทฏ Ex 13.4, 9 (b) Find P (X < 2), P (X โค 2), P(X โฅ 2). Our probability distribution table is P(X < 2) = P(X = 0) + P(X = 1) = k + 2k = 3k = 3 ร 1๏ทฎ6๏ทฏ = ๐๏ทฎ๐๏ทฏ P(X โค 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 3k = 6k = 6 ร 1๏ทฎ6๏ทฏ = 1 P(X โฅ 2) = P(X = 2) + P(x = 3) + P(X = 4) + โฆโฆ = 3k + 0 + 0 + โฆโฆโฆ. = 3k = 3 ร 1๏ทฎ6๏ทฏ = ๐๏ทฎ๐๏ทฏ