# Ex 13.4, 10 - Chapter 13 Class 12 Probability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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

Chapter 13 Class 12 Probability

Serial order wise

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