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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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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) = 𝑘, 𝑖𝑓 𝑥=0﷮2𝑘, 𝑖𝑓 𝑥=1﷮3𝑘, 𝑖𝑓 𝑥=2﷮0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒﷯﷯ (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﷮4﷮𝑃(𝑋)﷯ = 1 P(x = 0) + P(x = 1) + P(x = 2) = 1 k + 2k + 3k = 1 6k = 1 k = 𝟏﷮𝟔﷯ Ex 13.4, 9 (b) Find P (X < 2), P (X ≤ 2), P(X ≥ 2). Our probability distribution table is P(X < 2) = P(X = 0) + P(X = 1) = k + 2k = 3k = 3 × 1﷮6﷯ = 𝟏﷮𝟐﷯ P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 3k = 6k = 6 × 1﷮6﷯ = 1 P(X ≥ 2) = P(X = 2) + P(x = 3) + P(X = 4) + …… = 3k + 0 + 0 + ………. = 3k = 3 × 1﷮6﷯ = 𝟏﷮𝟐﷯

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.