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Chapter 13 Class 12 Probability

Misc 6 - In a hurdle race, a player has to cross 10 hurdles

Misc 6 - Chapter 13 Class 12 Probability - Part 2
Misc 6 - Chapter 13 Class 12 Probability - Part 3 Misc 6 - Chapter 13 Class 12 Probability - Part 4

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Transcript

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^(π’βˆ’π’™) 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/πŸ”)^𝒙 (πŸ“/πŸ”)^(𝟏𝟎 βˆ’ 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 βˆ’0) + 10C1(1/6)^1 (5/6)^(10 βˆ’ 1) = 1 Γ— 1 Γ— (5/6)^10 + 10 Γ— 1/6 Γ— (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 = πŸ“^𝟏𝟎/(𝟐 Γ— πŸ”^πŸ— )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.