Question

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 1

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) = nCxqn - xpx, x = 0, 1, 2, ...., n

P(X = r) = 10Cr(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)