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Ex 13.5

Ex 13.5, 1
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Ex 13.5, 2 Deleted for CBSE Board 2023 Exams

Ex 13.5, 3 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 4 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 5 Deleted for CBSE Board 2023 Exams

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Ex 13.5, 7 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 8 Deleted for CBSE Board 2023 Exams

Ex 13.5, 9 Deleted for CBSE Board 2023 Exams

Ex 13.5, 10 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 11 Deleted for CBSE Board 2023 Exams

Ex 13.5, 12 Deleted for CBSE Board 2023 Exams

Ex 13.5, 13 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at Feb. 15, 2020 by Teachoo

Ex 13.5, 6 A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?If 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 marked on bulb drawn Picking balls is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π Here, n = number of times we pick the bulb = 4 p = Probability of getting digit 0 = 1/10 q = 1 β p = 1 β 1/10 = 9/10 Hence, P(X = x) = 4Cx (π/ππ)^π (π/ππ)^(πβπ) We need to find Probability that none is marked with 0 i.e. P(X = 0) P(X = 0) = 4C0(1/10)^0 (9/10)^(4 β0) = (4 !)/((4 β 0) ! 0 !) Γ1Γ(9/10)^4 = (4 !)/(4 !)Γγ9 γ^4/γ10γ^( 4) = γ9 γ^4/γ10γ^( 4) = (π/ππ)^π