In a hurdle race, a player has to cross 10 hurdles. The probability th

1.	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?
Answer» It is a case of Bernoulli trials, where success is crossing a hurdle successfully without knocking it down and n = 10. p = P(success) = 5/6 q = 1 - p = 1 - 5/6 = 1/6 Let X be the random variable that represents the number of times the player will knock down the hurdle. Clearly, X has a binomial distribution with n = 10 and p = 5/6 Therefore P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...., n P(X = r) = ¹⁰C_r(1/6)^r(5/6)^10-r P (player knocking down less than 2 hurdles) = P( x < 2) = P(0) + P(1) = ¹⁰C₀p⁰q¹⁰ + ¹⁰C₁p¹q⁹ = (5/6)¹⁰ + 10(1/6)¹(5/6)⁹ = (5/6)⁹[5/6 + 10/6] = (5/6)⁹ x 15/6 = 5/2 x (5/6)⁹ = 5¹⁰/(2 x 69)

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?

Answer»

It is a case of Bernoulli trials, where success is crossing a hurdle successfully without knocking it down and n = 10.

p = P(success) = 5/6

q = 1 - p = 1 - 5/6 = 1/6

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Clearly, X has a binomial distribution with n = 10 and p = 5/6

Therefore P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...., n

P(X = r) = ¹⁰C_r(1/6)^r(5/6)^10-r

P (player knocking down less than 2 hurdles) = P( x < 2)

= P(0) + P(1) = ¹⁰C₀p⁰q¹⁰ + ¹⁰C₁p¹q⁹

= (5/6)¹⁰ + 10(1/6)¹(5/6)⁹ = (5/6)⁹[5/6 + 10/6] = (5/6)⁹ x 15/6 = 5/2 x (5/6)⁹ = 5¹⁰/(2 x 69)

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