# Example 28 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

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) = ∑2_(𝒊 = 𝟏)^𝒏▒𝒙𝒊𝒑𝒊 = 1 × 1/6+2 × 1/6+ 3 × 1/6+ 3 × 1/6+ 5 × 1/6+ 6 × 1/6 = 21/6 The variance of x is given by : Var (𝑿)=𝑬(𝑿^𝟐 )−[𝑬(𝑿)]^𝟐 So, finding 𝐸(𝑋^2 ) E(𝑋^2 )=∑2_(𝑖 = 1)^𝑛▒〖〖𝑥_𝑖〗^2 𝑝𝑖〗 = 12 × 1/6+22 × 1/6+ 32 × 1/6+ 42 × 1/6+ 52 × 1/6+ 62 × 1/6 = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6 Now, Var (𝑋)=𝐸(𝑋^2 )−[𝐸(𝑋)]^2 = 91/6−[21/6]^2 = 91/6−441/36 = (546 − 441)/36 = 105/36 = 35/12 Hence, variance is 𝟑𝟓/𝟏𝟐

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Chapter 13 Class 12 Probability (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.