auto ad

Friday, December 30, 2016

hurdle problem from miscellaneous cbse ncert mathematics probability

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 the player knocks down out of 10 hurdles

Assume X follows binomial distribution with
n=10

p = 1-(5/6) [because we defined X in terms of the hurdles knocked down ]
p=(1 /6)
q = 1 -p
q =(5/6)

P[X=r] = nCr prq(n-r) , r = 0,1,2,...,n

P[X=r] = 10Cr (1/6)r(5/6)(10 - r) , r = 0,1,2,...,10

P[the player will knock down fewer than two hurdles] = P[ X < 2 ]

P[ X < 2 ] = P[X=0] + P[X=1]

No comments:

Post a Comment

please leave your comments