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Algebra

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7 Jan 2009 10:48:57 IST

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423

a fair die is thrown 6 times...............

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A fair dice is thrown six times and the list of numbers showing up is noted.The probability that among the numbers 1 to 6 only 4 numbers appear in the list is?

(a)64 / 729

(b)165 / 729

(c)165 / 1944

(d)325 / 648

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Comments (3)

Sagar Saxena

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24 Mar 2009 02:20:46 IST

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1-(6c2*(2/6)^6 - 2c1*(1/6)^6)

solve dis to get eqtn

1-probability of gettting exactly 2 no. in d list

Vikram Pandit

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24 Mar 2009 10:55:22 IST

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Only a mere assumption....isn't the answer 64/729???

Kevin Arnold

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24 Mar 2009 19:41:56 IST

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isit D)

u can do dis in 2 ways...1-P(1or 2 numbers appearing) which i guess has been done by sir..

so i shall tell the 2nd method..

we can select those 4 numbers in ⁶C₄ ways...now 4 of the 6 tries are gonna contain these 4 numbers..the rest of the 2 tries also contain 2 of

the same 4 numbers...these can be 2 similar numbers or 2 different numbers...so we have 2 cases 1) 2 similar numbers, 2) different numbers

for 1) we have ⁶C₄ * ⁴C₁ *6!/3!...

for 2) we have ⁶C₄ * ⁴C₂ *6!/(2!)²

so event space= 1) +2)

sample space is 6⁶

this way i guess the answer comes as D) if there has not been any agonising silly mistake on my part..

cheers!!!

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