# Ex 13.4, 13 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

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

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 You are here

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

Ex 13.4, 13 Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.Ex 13.4, 13 Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.Thus, Probability distribution is Now, we need to find variance Var (𝑋) "= " 𝐸(𝑋^2 )−[𝐸(𝑋)]^2 Finding E(X) E(X) = ∑_(𝑖 = 1)^𝑛▒𝑥𝑖𝑝𝑖 = 2 ×1/36+3 × 2/36+4 × 3/36+5 ×4/36+6 × 5/36+7 × 6/36 +8 × 5/36+9 × 4/36+10 ×3/36+11 × 2/36+12 × 1/36 = (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)/36 = (252 )/36 = 7 Thus E(x) = 7 Finding E(𝑿^𝟐 ) E(𝑿^𝟐 )=∑_(𝑖=1)^𝑛▒〖𝑥^2 𝑖𝑃𝑖〗 = 2^2×1/36 +3^2×2/36 +4^2 ×3/36+5^2×4/36+6^2×5/36+7^2× 6/36 +8^2 ×5/36+9^2× 4/36+〖10〗^2×3/36+〖11〗^2 ×2/36+〖12〗^2×1/36 = (4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144)/36 = 1974/36 = 𝟑𝟐𝟗/𝟔 Now, Var (𝑿) "= " 𝑬(𝑿^𝟐 )−[𝑬(𝑿)]^𝟐 = 329/6−(7)^2 = (329 − 6 × 49)/6 = 35/6 = 5.833 Standard deviation is given by 𝝈_𝒙=√(𝒗𝒂𝒓 (𝑿) ) =√(35/6) =√5.833 = 2.415 43 × 3 = 129 44 × 4 = 176 45 × 5 = 225 4824 × 4 = 19296 4825 × 5 = 24125 4826 × 6 = 28956