Question 7 - Chapter 13 Class 12 Probability (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 14 Important
Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important
Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important You are here
Question 8 Important
Question 3 Important
Question 6 Important
Question 11 Important
Question 15
Question 10 Important
Question 11 Important
Question 4 Important
Question 6 Important
Question 10 Important
Question 13 Important
Question 13
Example 23 Important
Question 2 Important
Question 4
Question 6 Important
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
Question 7 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