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

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Misc 6 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﷯ n = number of hurdles = 10 Given, Probability that he will clear hurdle = 5﷮6﷯ ⇒ q = 5﷮6﷯ 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﷯﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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