# Misc 6

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc 6 In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 56 . 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 𝒒𝒏−𝒙 𝒑𝒙 n = number of hurdles = 10 Given, Probability that he will clear hurdle = 56 ⇒ q = 56 p = 1 – q = 1 – 56 = 16 Hence, P(X = x) = 10Cx 𝟏𝟔𝒙 𝟓𝟔𝟏𝟎 − 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 160 5610 −0 + 10C1 161 5610 − 1 = 1 × 1 × 5610 + 10 × 16 × 569 = 5610+ 106 569 = 56 569+ 106 569 = 56 569+ 106 569 = 56+ 106 569 = 156 569 = 52 569 = 𝟓𝟏𝟎𝟐 × 𝟔𝟗

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Misc 6 Important You are here

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Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

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.