Ex 13.4, 10 - Find mean number of heads in three tosses - Ex 13.4

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  1. Chapter 13 Class 12 Probability
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Ex 13.4, 10 (Method 1) Find the mean number of heads in three tosses of a fair coin. Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx Here, n = number of coins tosses = 3 p = Probability of heads = 1 2 q = 1 p = 1 1 2 = 1 2 Hence, P(X = x) = 3Cx 1 2 1 2 3 P(X = x) = 3Cx 1 2 3 + P(X = x) = 3Cx We can get 0 heads, 1 heads, 2 heads, 3 heads So, values of X can be 0, 1, 2, 3 Hence, P(X = 0) = 3C0 1 2 4 = 1 1 2 3 = 1 8 P(X = 1) = 3C1 1 2 4 = 3 1 2 3 = 3 8 P(X = 2) = 3C2 1 2 4 = 3 1 2 3 = 3 8 P(X = 3) = 3C3 1 2 4 = 1 1 2 3 = 1 8 So the Probability distribution The mean number is given by = = = 1 = 0 1 8 +1 3 8 + 2 3 8 + 3 1 8 = 0 + 3 8 + 6 8 + 3 8 = 12 8 = 3 2 = 1.5 Ex 13.4, 10 (Method 2) Find the mean number of heads in three tosses of a fair coin. Let X : Number of heads We toss 3 coins simultaneously So, we can get 0 heads, 1 heads , 2 heads or 3 heads. So, value of X can be 0, 1, 2, 3 So the Probability distribution The mean number is given by = = = 1 = 0 1 8 +1 3 8 + 2 3 8 + 3 1 8 = 0 + 3 8 + 6 8 + 3 8 = 12 8 = 3 2

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