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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 You are here

Ex 13.4, 9

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 May 29, 2018 by Teachoo

Ex 13.4, 8 A random variable X has the following probability distribution: Determine (i) k Since X is a random variable , its Sum of Probabilities is equal to 1 07𝑃(𝑋) = 1 P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1 9k + 10k2 = 1 10k2 + 9k – 1 = 0 10k2 + 10k – k – 1 = 0 10k2 + 10k – k – 1 = 0 10k(k + 1) – 1(k + 1) = 0 (10k – 1)(k + 1) = 0 So, k = 110 & k = –1 But k is probability, So, it cannot be negative, Hence, k = 𝟏𝟏𝟎 Ex 13.4, 8 A random variable X has the following probability distribution: Determine (ii) P(X < 3) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 2k = 3k = 3 × 110 = 𝟑𝟏𝟎 Ex 13.4, 8 A random variable X has the following probability distribution: Determine (iii) P(X > 6) P(X > 6) = P(X = 7) = 7k2 + k = 7 1102+ 110 = 7 × 1100+ 110 = 7100+ 110 = 𝟏𝟕𝟏𝟎𝟎 Ex 13.4, 8 A random variable X has the following probability distribution: Determine (iv) P(0 < X < 3) P(0 < X < 3) = P(X = 1) + P(X = 2) = k + 2k = 3k = 3 × 110 = 𝟑𝟏𝟎