Question 13 - Chapter 13 Class 12 Probability (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
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Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
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Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
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Question 13 Deleted for CBSE Board 2024 Exams You are here
Example 23 Important
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Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
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^(𝒏−𝒙) 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 (𝟗/𝟏𝟎)^(𝟏𝟐−𝒙) Picking articles is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 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 (𝟗/𝟏𝟎)^(𝟏𝟐−𝒙) 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 − 9) = (12 !)/((12 − 9) ! 9 !) (1/10)^9 (9/10)^3 = (12 × 11 × 10 × 9 !)/(3 ! 9 !) \ ×1/〖10〗^9 (9/10)^3 = (12 × 11 × 10)/(3 × 2 × 1) × 1/〖10〗^9 (9/10)^3= (2 × 11)/〖10〗^8 9^3/〖10〗^3 = 22 (𝟗^𝟑/〖𝟏𝟎〗^𝟏𝟏 )