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Chapter 13 Class 12 Probability

Ex 13.5, 13 - Known that 10% articles manufactured are defective

Ex 13.5, 13 - Chapter 13 Class 12 Probability - Part 2
Ex 13.5, 13 - Chapter 13 Class 12 Probability - Part 3

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Question 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 – 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 𝒒^(π’βˆ’π’™) 𝒑^𝒙 Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 – p = 1 – 1/10 = 9/10 Hence, P(X = x) = 12Cx (𝟏/𝟏𝟎)^𝒙 (πŸ—/𝟏𝟎)^(πŸπŸβˆ’π’™) Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(π’βˆ’π’™) 𝒑^𝒙 Here, n = number of articles = 12 p = Probability of getting defective articles = 10% = 10/100 = 1/10 q = 1 – p = 1 – 1/10 = 9/10 Hence, P(X = x) = 12Cx (𝟏/𝟏𝟎)^𝒙 (πŸ—/𝟏𝟎)^(πŸπŸβˆ’π’™) 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 βˆ’ 9) = (12 !)/((12 βˆ’ 9) ! 9 !) (1/10)^9 (9/10)^3 = (12 Γ— 11 Γ— 10 Γ— 9 !)/(3 ! 9 !) \ Γ—1/γ€–10γ€—^9 (9/10)^3 = (12 Γ— 11 Γ— 10)/(3 Γ— 2 Γ— 1) Γ— 1/γ€–10γ€—^9 (9/10)^3= (2 Γ— 11)/γ€–10γ€—^8 9^3/γ€–10γ€—^3 = 22 (πŸ—^πŸ‘/γ€–πŸπŸŽγ€—^𝟏𝟏 )

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