Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Example 28 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Examples
Example 1
Example 2
Example 3
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8
Example 9 Important
Example 10
Example 11 Important
Example 12 Important
Example 13 Important
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important
Example 20 Important
Example 21 Important
Example 22
Example 23
Example 24 Important
Example 25 Important
Example 26 Important
Example 27
Example 28 Important You are here
Example 29 Important
Example 30 Important
Example 31 Important
Example 32 Important
Example 33 Important
Example 34
Example 35
Example 36 Important
Example 37 Important
Chapter 13 Class 12 Probability
Serial order wise
Transcript
Example 28 Find the variance of the number obtained on a throw of an unbiased die. Let X be number obtained on a throw So, value of X can be 1, 2, 3, 4, 5 or 6 Since die unbiased, Probability of getting of each number is equal P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = 1/6 Hence, probability distribution The mean Expectation value is given by E(X) = β2_(π = π)^πβππππ = 1 Γ 1/6+2 Γ 1/6+ 3 Γ 1/6+ 3 Γ 1/6+ 5 Γ 1/6+ 6 Γ 1/6 = 21/6 The variance of x is given by : Var (πΏ)=π¬(πΏ^π )β[π¬(πΏ)]^π So, finding πΈ(π^2 ) E(π^2 )=β2_(π = 1)^πβγγπ₯_πγ^2 ππγ = 12 Γ 1/6+22 Γ 1/6+ 32 Γ 1/6+ 42 Γ 1/6+ 52 Γ 1/6+ 62 Γ 1/6 = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6 Now, Var (π)=πΈ(π^2 )β[πΈ(π)]^2 = 91/6β[21/6]^2 = 91/6β441/36 = (546 β 441)/36 = 105/36 = 35/12 Hence, variance is ππ/ππ
