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Suppose 3 dice are rolled. the probability that sum of the numbers of the dice is 10 is ____________
asked in Mathematical Logic by Boss (8k points) 1 13 94 | 87 views

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For this we have to find number of ways we can get a sum of 10 in 3 rolls of dice which is nothing but the number of positive integral solutions of the equation :

x1 + x2 + x3  =  10  where  1 <= x1,x2,x3 <= 6

Now number of possible integral solutions can be found as :

                        Coefficient of x10  in  (x + x2 + ........ + x6)3 

             ==>     Coefficient of x10  in   x3 (1 + x +.....+ x5)3

             ==>     Coefficient of x7  in    (1 + x +.....+ x5)3

             ==>     Coefficient of x7  in    (1 - x6)3  . (1 - x)-3

Now only relevant terms will be extracted from (1 - x6)3  and accordingly powers of (1 - x)-3 will be taken to get the coefficients    of   x7 .

           So    (1 - x6)3     =  (1 - 3x6 + 3x12  -  x18) out of which only first 2 terms are relevant..

So  , due to the term "1"  , coefficient of x7 is needed in (1 - x)-3 which is :   3-1+7C7  =  9C7  =   36

        due to term "x" ,  coefficient of x is needed in (1 - x)-3 which is  :  3-1+1C1     =   3C1 =   3

    So due to -3x6 , we will have coefficient   =   -3 * 3  =  -9

    Hence net coefficient of x7     =   36  -  9    =     27

    Hence P(sum of 3 dices is 10)    =   n(sum of 3 dices is10) / Total outcome

                                                    =   27 / (6 * 6 * 6)

                                                    =   27 / 216

                                                   =   1 / 8

answered by Veteran (87k points) 15 57 292
edited by
@habib your approach is fine , but answer is incorrect it will be $\frac{27}{216}$
Awesome answer. (Y)

After taking $x^{3}$ common, you are finding the coefficient of $x^{7}$. But in the end, you again find the coefficient of $x^{10}$.

In the whole calculation you are deriving coefficients for  for x10  I think should be for x7 .

(1 * 2+7C7 ) + (-3 2+1C1 )

36-9

27/216

 

1. 1,3,6 (3!) =6

2. 1,4,5(3!)=6

3. 2,2,6(3!/2!=3)

4. 2,3,5(3!)=6

5. 2,4,4(3!/2!=3)

6.3,3,4(3!/2!=3) 

i.e 18+9= 27/216...

Another silly mistake by me..Thank u all for rectifying me..
any video or links to understand the concept used in ur explanation please

am finding it a little difficult to understand

you may search for "Generating Functions"

one such video is

thank you :) the video was helpful

  So    (1 - x6)3     =  (1 - 3x6 + 3x12  -  x18) out of which only first 2 terms are relevant.. \\how is this expansion

So  , due to the term "1"  , coefficient of x7 is needed in (1 - x)-3 which is :   3-1+7C7  =  9C7  =   36

        due to term "x" ,  coefficient of x is needed in (1 - x)-3 which is  :  3-1+1C1     =   3C1 =   3

    So due to -3x6 , we will have coefficient   =   -3 * 3  =  -9

    Hence net coefficient of x7     =   36  -  9    =     27

    Hence P(sum of 3 dices is 10)    =   n(sum of 3 dices is10) / Total outcome

                                                    =   27 / (6 * 6 * 6)

                                                    =   27 / 216

                                                   =   1 / 8

and can someone help me only with this part

in the video...the audio part is earlier than the visual so couldnt understand the last part

  So    (1 - x6)3     =  (1 - 3x6 + 3x12  -  x18) out of which only first 2 terms are relevant.. \\how is this expansion

Binomial Expansion.

So  , due to the term "1"  , coefficient of x7 is needed in (1 - x)-3 which is :   3-1+7C7  =  9C7  =   36

        due to term "x" ,  coefficient of x is needed in (1 - x)-3 which is  :  3-1+1C1     =   3C1 =   3

    So due to -3x6 , we will have coefficient   =   -3 * 3  =  -9

    Hence net coefficient of x7     =   36  -  9    =     27

 

this part too please



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