# Ex 13.4, 11 - Chapter 13 Class 12 Probability

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 13.4, 11(Method 1) Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X. Tossing a Die is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 Here, n = number of tosses of die = 2 p = Probability six appears = 16 q = 1 – p = 1 – 16 = 56 Hence, ⇒ P(X = x) = 2Cx 𝟏𝟔𝒙 𝟓𝟔𝟐 − 𝒙 Since pair of die are thrown, We can get 0 six, 1 six or 2 six So, X can be 0, 1 , 2 Putting values in (1) P(X = 0) = 2C0 160 562 −0 = 1 × 562 = 2536 P(X = 1) = 2C1 161 562 − 1 = 2 × 16 × 56 = 1036 P(X = 2) = 2C2 162 562 − 2 = 1 × 162 = 136 Thus, probability distribution is The Expectation of X is given by E(X) = 𝜇= 𝑖=1𝑛𝑥𝑖𝑝𝑖 = 0 × 2536+1 × 1036+2 × 136 = 1036+ 236 = 1236 = 𝟏𝟑 Ex 13.4, 11(Method 2) Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X. Let X be the number of sixes occur So, value of X can be 0, 1 or 2 Total number of possible outcomes = 36 S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) So, our probability distribution is The Expectation of X is given by E(X) = 𝜇= 𝑖=1𝑛𝑥𝑖𝑝𝑖 = 0 × 2536+1 × 1036+2 × 136 = 1036+ 236 = 1236 = 𝟏𝟑

Chapter 13 Class 12 Probability

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.