# Ex 13.4, 4

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 13.4, 4 Find the probability distribution of (i) number of heads in two tosses of a coin. Let X: Number of heads We toss coin twice So, we can get 0 heads, 1 heads or 2 heads. So, value of X can be 0, 1, 2 So the Probability distribution Ex 13.4, 4 (Method 1) Find the probability distribution of (ii) number of tails in the simultaneous tosses of three coins. Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 Here, n = number of coins tosses = 3 p = Probability of tail = 12 q = 1 – p = 1 – 12 = 12 Hence, ⇒ P(X = x) = 3Cx 12𝑥 123 − 𝑥 P(X = x) = 3Cx 123 − 𝑥 + 𝑥 P(X = x) = 3Cx 𝟏𝟐𝟑 We can get 0 tails, 1 tail , 2 tails, 3 tails So, values of X can be 0, 1, 2, 3 Hence, P(X = 0) = 3C0 124 = 1 × 123 = 18 P(X = 1) = 3C1 124 = 3 × 123 = 38 P(X = 2) = 3C2 124 = 3 × 123 = 38 P(X = 3) = 3C3 124 = 1 × 123 = 18 So, probability distribution is Ex 13.4, 4(Method 2) Find the probability distribution of (ii) number of tails in the simultaneous tosses of three coins. Let X : Number of tails We toss 3 coins simultaneously So, we can get 0 tails, 1 tails , 2 tails or 3 tails. So, value of X can be 0, 1, 2, 3 So the Probability distribution Ex 13.4, 4 (Method 1) Find the probability distribution of (iii) number of heads in four tosses of a coin. Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 Here, n = number of coins tosses = 4 p = Probability of head = 12 q = 1 – p = 1 – 12 = 12 Hence, ⇒ P(X = x) = 4Cx 12𝑥 124 − 𝑥 P(X = x) = 4Cx 124 − 𝑥 + 𝑥 P(X = x) = 4Cx 𝟏𝟐𝟒 We can get 0 heads, 1 heads , 2 heads , 3 heads , 4 heads So, values of X can be 0, 1, 2, 3, 4 Hence, P(X = 0) = 4C0 124 = 1 × 124 = 116 P(X = 1) = 4C1 124 = 4 × 124 = 416 = 14 P(X = 2) = 4C2 124 = 6 × 124 = 616 = 38 P(X = 3) = 4C3 124 = 4 × 124 = 416 = 14 P(X = 4) = 4C4 124 = 1 × 124 = 116 So, probability distribution is Ex 13.4, 4 (Method 2) Find the probability distribution of (iii) number of heads in four tosses of a coin. Let X : Number of heads When toss 4 coins simultaneously, we can get So, we can get 0 heads, 1 heads , 2 heads , 3 heads , 4 heads So, values of X can be 0, 1, 2, 3, 4 So, our probability distribution is

Chapter 13 Class 12 Probability

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .