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Ex 13.4, 4 - Find probability distribution of (i) number of - Ex 13.4

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  1. Chapter 13 Class 12 Probability
  2. Serial order wise
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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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ Here, n = number of coins tosses = 3 p = Probability of tail = 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 tails, 1 tail , 2 tails, 3 tails 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, 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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ Here, n = number of coins tosses = 4 p = Probability of head = 1﷮2﷯ q = 1 – p = 1 – 1﷮2﷯ = 1﷮2﷯ Hence, ⇒ P(X = x) = 4Cx 1﷮2﷯﷯﷮𝑥﷯ 1﷮2﷯﷯﷮4 − 𝑥﷯ P(X = x) = 4Cx 1﷮2﷯﷯﷮4 − 𝑥 + 𝑥﷯ P(X = x) = 4Cx 𝟏﷮𝟐﷯﷯﷮𝟒﷯ 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 1﷮2﷯﷯﷮4﷯ = 1 × 1﷮2﷯﷯﷮4﷯ = 1﷮16﷯ P(X = 1) = 4C1 1﷮2﷯﷯﷮4﷯ = 4 × 1﷮2﷯﷯﷮4﷯ = 4﷮16﷯ = 1﷮4﷯ P(X = 2) = 4C2 1﷮2﷯﷯﷮4﷯ = 6 × 1﷮2﷯﷯﷮4﷯ = 6﷮16﷯ = 3﷮8﷯ P(X = 3) = 4C3 1﷮2﷯﷯﷮4﷯ = 4 × 1﷮2﷯﷯﷮4﷯ = 4﷮16﷯ = 1﷮4﷯ P(X = 4) = 4C4 1﷮2﷯﷯﷮4﷯ = 1 × 1﷮2﷯﷯﷮4﷯ = 1﷮16﷯ 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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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