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Ex 13.4, 9 - Random variable X has probability distribution - Ex 13.4

Ex 13.4, 9 - Chapter 13 Class 12 Probability - Part 2
Ex 13.4, 9 - Chapter 13 Class 12 Probability - Part 3
Ex 13.4, 9 - Chapter 13 Class 12 Probability - Part 4

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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‘˜, ๐‘–๐‘“ ๐‘ฅ=0๏ทฎ2๐‘˜, ๐‘–๐‘“ ๐‘ฅ=1๏ทฎ3๐‘˜, ๐‘–๐‘“ ๐‘ฅ=2๏ทฎ0, ๐‘œ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘ค๐‘–๐‘ 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‘ƒ(๐‘‹)๏ทฏ = 1 P(x = 0) + P(x = 1) + P(x = 2) = 1 k + 2k + 3k = 1 6k = 1 k = ๐Ÿ๏ทฎ๐Ÿ”๏ทฏ Ex 13.4, 9 (b) Find P (X < 2), P (X โ‰ค 2), P(X โ‰ฅ 2). Our probability distribution table is P(X < 2) = P(X = 0) + P(X = 1) = k + 2k = 3k = 3 ร— 1๏ทฎ6๏ทฏ = ๐Ÿ๏ทฎ๐Ÿ๏ทฏ P(X โ‰ค 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 3k = 6k = 6 ร— 1๏ทฎ6๏ทฏ = 1 P(X โ‰ฅ 2) = P(X = 2) + P(x = 3) + P(X = 4) + โ€ฆโ€ฆ = 3k + 0 + 0 + โ€ฆโ€ฆโ€ฆ. = 3k = 3 ร— 1๏ทฎ6๏ทฏ = ๐Ÿ๏ทฎ๐Ÿ๏ทฏ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.