Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Binomial Distribution
Question 12 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams You are here
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 14 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 15 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams You are here
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Misc 4
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Binomial Distribution
Last updated at May 29, 2023 by Teachoo
Question 2 In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6 . What is the probability that he will knock down fewer than 2 hurdles?Let X : be the number of hurdles that player knocks down Crossing a hurdle is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of hurdles = 10 Given, Probability that he will clear hurdle = 5/6 So, q = 5/6 Thus, p = 1 – q = 1 – 5/6 = 1/6 Hence, P(X = x) = 10Cx (𝟏/𝟔)^𝒙 (𝟓/𝟔)^(𝟏𝟎 − 𝒙) We need to find Probability that he will knock down fewer than 2 hurdles P(he will knock down fewer than 2 hurdles) = P(knock 0 hurdles) + P(knock 1 hurdles) = P(X = 0) + P(X = 1) = 10C0(1/6)^0 (5/6)^(10 −0) + 10C1(1/6)^1 (5/6)^(10 − 1) = 1 × 1 × (5/6)^10 + 10 × 1/6 × (5/6)^9 = (5/6)^10+10/6 (5/6)^9 = 5/6 (5/6)^9+10/6 (5/6)^9 = 5/6 (5/6)^9+10/6 (5/6)^9 = (5/6+10/6) (5/6)^9 = 15/6 (5/6)^9 = 5/2 (5/6)^9 = 𝟓^𝟏𝟎/(𝟐 × 𝟔^𝟗 )