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:
Let p and q are the probabilities that player will clear and knock down the hurdle.
Given p = 5/6
So q = 1 - 5/6
=> q = 1/6
Let X is the random variable that represents the number of times the player will knock down the hurdle.
So by Binomial distribution
P(X =x) = nCx * pn-x * qx
Now probability that player knocking down less that 2 hurdle = P(X < 2) = P(X = 0) + P(X = 1)
= 10C0 * q0 * p10 + 10C1 * q1 * p9
= 10C0 * (5/6)10 + 10C1 * (1/6)1 * (5/6)9
= (5/6)9 * (5/6 + 10/6)
= (5/6)9 *15/6
= (5/2)(5/6)9
= 510 /(2* 69 )
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