IIT JEE , AIEEE Forum - Mathematics , Algebra - the probability of getting 10 in a single throw of 3 fair dice is

amit198

the probability of getting 10 in a single throw of 3 fair dice is

spideyunlimited

10 can be expressed in sums of three as :

6,3,1 .................rearranged in 3! ways
6,2,2 .................rearranged in 3! /2 ways
5,4,1 .................rearranged in 3! ways
5,3,2 .................rearranged in 3! ways
4,3,3.................rearranged in 3!/2 ways

2,4,4.................rearranged in 3!/2 ways

so favourable cases = 3 * 3! + 3 * 3!/2

= 18 + 9 = 27

total cases = 6*6*6 = 216

So probability = 27 / 216

sweet was correct :)

(Edited)

sweet08

is ans 1/8

ramkumar_november

the no. of favourable cases must be

the coefficient of x¹⁰ in (x+x²+x³+x⁴+x⁵+x⁶)³

toatal no. of cases is 6³ = 216

sweet08

2,3,5---------3!

2,4,4---------3!/2

2,2,6------------3!/2

3,3,4-----------3!/2

3,6,1-----------3!

4,5,1----------3!

so=27/216=1/8

spideyunlimited

sweet, you don't have to divide by 2 ...

both ave equal chances of getting the common number,,, we don't consder it as an unnecesry repitition.

amit198

@ sweet 08 you are correct tell me how

spideyunlimited

OH my, i forgot 4,3,3

sorry my solultion is wrong..
error regretted

spideyunlimited

correct now! corrected my solution .

joyfrancis

RAGTE!!!..2 case kha gaya tu..

442 and 433 ...

favourable are 27 ..

joyfrancis

shabaash!!..pakad lee galti :)..badiya..!!

spideyunlimited

abbe i corrected mere bhai :P

joyfrancis

haan haan..good good ..

RyuAmakusa

SORRY

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