# Ex 13.5, 13 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Ex 13.5

Ex 13.5, 1
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Chapter 13 Class 12 Probability (Term 2)

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Ex 13.5, 13 It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective? In each of the following, choose the correct answer:vIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i.e. Probability success = P then Probability failure = q = 1 β P (4) Probability of success (p) is same for all trials Let X : be the number of defective articles Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 β p = 1 β 1/10 = 9/10 Hence, P(X = x) = 12Cx (π/ππ)^π (π/ππ)^(ππβπ) Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 β p = 1 β 1/10 = 9/10 Hence, P(X = x) = 12Cx (π/ππ)^π (π/ππ)^(ππβπ) We need to find Probability of getting 9 defective articles out of 12 i.e. P(X = 9) P(X = 9) = 12C9 (1/10)^9 (9/10)^(12 β 9) = (12 !)/((12 β 9) ! 9 !) (1/10)^9 (9/10)^3 = (12 Γ 11 Γ 10 Γ 9 !)/(3 ! 9 !) \ Γ1/γ10γ^9 (9/10)^3 = (12 Γ 11 Γ 10)/(3 Γ 2 Γ 1) Γ 1/γ10γ^9 (9/10)^3= (2 Γ 11)/γ10γ^8 9^3/γ10γ^3 = 22 (π^π/γππγ^ππ )