Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Probability Distribution
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 (i) Deleted for CBSE Board 2024 Exams
Question 4 (ii) Deleted for CBSE Board 2024 Exams
Question 4 (iii) Important Deleted for CBSE Board 2024 Exams
Question 5 (i) Important Deleted for CBSE Board 2024 Exams
Question 5 (ii) Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams You are here
Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Question 15 Deleted for CBSE Board 2024 Exams
Question 16 (MCQ) Deleted for CBSE Board 2024 Exams
Question 17 (MCQ) Important Deleted for CBSE Board 2024 Exams
Probability Distribution
Last updated at Aug. 9, 2023 by Teachoo
Question 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 Question 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