Misc 5 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Binomial Distribution
Ex 13.5, 12 Deleted for CBSE Board 2022 Exams
Ex 13.5, 2 Deleted for CBSE Board 2022 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 9 Deleted for CBSE Board 2022 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 11 Deleted for CBSE Board 2022 Exams
Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
Example 32 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams
Misc 6 Important Deleted for CBSE Board 2022 Exams
Misc 7 Important
Ex 13.5, 7 Important Deleted for CBSE Board 2022 Exams
Example 31 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 5 Deleted for CBSE Board 2022 Exams
Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams
Misc 5 Important Deleted for CBSE Board 2022 Exams You are here
Misc 9 Deleted for CBSE Board 2022 Exams
Misc 4 Deleted for CBSE Board 2022 Exams
Misc 10 Important
Example 35 Deleted for CBSE Board 2022 Exams
Ex 13.5, 8 Deleted for CBSE Board 2022 Exams
Example 34 Deleted for CBSE Board 2022 Exams
Binomial Distribution
Misc 5 An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear 'X' mark. (ii) not more than 2 will bear 'Y' mark. (iii) at least one ball will bear 'Y' mark. (iv) the number of balls with 'X' mark and 'Y' mark will be equal.Let X : Number of balls with mark ‘X’ Drawing a ball is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of balls drawn = 6 p = Probability of getting ball with ‘X’ mark = 10/25 = 2/5 q = 1 – p = 1 – 2/5 = 3/5 Hence, P(X = x) = 6Cx (𝟐/𝟓)^𝒙 (𝟑/𝟓)^(𝟔 − 𝒙) Probability that all will bear 'X' mark. Probability all balls has ‘X’ mark = P(X = 6) Putting x = 6 in (1) P(X = 6) = 6C6 (2/5)^6 (3/5)^(6 −6) = 6C6 (2/5)^6 (3/5)^0 = 1 × (2/5)^6× 1 = (𝟐/𝟓)^𝟔 Probability that not more than 2 will bear 'Y' mark. P(not more than 2 bear Y) = P(6X, 0Y) + P(5X, 1Y) + P(4X, 2Y) = P(X = 6) + P(X = 5) + P(X = 4) = 6C6(2/5)^6 (3/5)^(6−6) "+ 6C5" (2/5)^5 (3/5)^(6−5)+"6C4" 〖 (2/5)〗^4 (3/5)^(6−4) = 6C6(2/5)^6 (3/5)^0 "+ 6C5" (2/5)^5 (3/5)^1+"6C4" 〖 (2/5)〗^4 (3/5)^2 = 1 × (2/5)^6 "×" (3/5)^0 "+ 6 ×" (2/5)^5 (3/5)+"15" 〖 (2/5)〗^4 (3/5)^2 = (2/4)^4 [(2/5)^2+"6 ×" (2/5)(3/5)+15(3/5)^2 ] = (2/4)^4 [4/25+36/25+135/25]=(2/5)^4 [175/25] =𝟕(𝟐/𝟓)^𝟒 (iii) Probability that at least one ball will bear 'Y' mark. P(atleast one bears ‘Y’) = 1 – P(no balls bear ‘Y’) = 1 – P(all ball bears ‘X’) = 1 – P(X = 6) = 1 – 6C6(2/5)^6 (3/5)^(6−6) = 1 – 6C6(2/5)^6 (3/5)^0 = 1 – 1 × (2/5)^6 × 1 = 1 – (𝟐/𝟓)^𝟔 (iv) Probability that number of balls with 'X' mark & 'Y' mark will be equal. So, we will have 3X & 3Y balls P(X & Y marked balls are equal) = P(X = 3) = 6C3 (2/5)^3 (3/5)^(6−3) = 6C3 (2/5)^3 (3/5)^3 = (6 × 5 × 4 × 3!)/(3! × 3 × 2 × 1)×8/125×27/125 = 5 × 4 × 8/125 × 27/125 = 𝟖𝟔𝟒/𝟑𝟏𝟐𝟓