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anpm_dev (111)

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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. (NCERT TBQ Probability Misc. Q-6) plz reply soon

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arpan1 (665)

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ans is (5/6)10 + 10C1(5/6)9 x (1/6)


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anpm_dev (111)

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oh i know the answer plz give the solution

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akhil_o (2709)

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prob to knock down 1 hurdle: 1/6

so in 10 jumps, prob to knock down only one hurdle:
10*1/6*(5/6)^9
for no hurdle knocked down:
(5/6)^10

so total P=10*1/6*(5/6)^9+(5/6)^10
=(5/6)^9(15/6)
=(5/6)^9(5/2)

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anpm_dev (111)

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is that a binomial trial

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YES IT IS

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akhil_o (2709)

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sorry...what is 'binomial trial'?
technical terms out of my reach!

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anpm_dev (111)

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plz solve this question for me too


a die is thrown again and again until three sixes are obtained. find the prob. of obtaining the third six in the sixth throw of the die.

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akhil_o (2709)

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ok see fix the 3rd six in the sixth throw..
this means that 2 sixes have already appeared in past 5 throws
so we have
5C2*(1/6)^3*(5/6)^3

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ANS. IS 5C2 x (1/6)2.(5/6)3.(1/6)

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computer001 (1849)

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u need to think only abt 1st 5 throws
so it'll be (1/6)^2*(5/6)^3*1/6
the last 1/6 accounts 4 the final throw
but getting 2 6s from 1st 5 is 5C2 ways
so v have 5C2(1/6)^2*(5/6)^3*1/6
EDITED:
seeing tht as usual i posted solution just after akhil did pls dun rate :D

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