# Ex 13.4, 9

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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

Chapter 13 Class 12 Probability

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .