Ex 13.4, 10

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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ 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﷯﷯﷮𝑥﷯ 1﷮2﷯﷯﷮3 − 𝑥﷯ P(X = x) = 3Cx 1﷮2﷯﷯﷮3 − 𝑥 + 𝑥﷯ P(X = x) = 3Cx 𝟏﷮𝟐﷯﷯﷮𝟑﷯ 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 𝝁=𝑬 𝑿﷯= 𝑖 = 1﷮𝑛﷮𝑥𝑖𝑝𝑖﷯ = 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 𝜇=𝐸 𝑥﷯= 𝑖 = 1﷮𝑛﷮𝑥𝑖𝑝𝑖﷯ = 0 × 1﷮8﷯+1 × 3﷮8﷯+ 2 × 3﷮8﷯+ 3 × 1﷮8﷯ = 0 + 3﷮8﷯+ 6﷮8﷯+ 3﷮8﷯ = 12﷮8﷯ = 3﷮2﷯