## Flipkart Interview Question

• 0

Country: India

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18
of 18 vote

three 0s and three 6s.

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0

Yes thats what I told :). Good

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0

Can you explain question little bit more ? I could not get what does it mean "Sum of faces is evenly distributed between 1 to 12".

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0

Well imagine case of 2 normal dice. The sums will be between 2-12 when you throw them. Also 2 will come less times ( 1 + 1) while 7 would come more times ( 4 + 3 , 3 + 4, 5 + 2, 2+5), therefore this is not even distribution.

You have to make sure that the probability of each number between 1-12 is equal. Having 3 faces 0 and 3 faces 6 does that. Calculate by yourself

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0

36 possibilities and 12 possible values of sum, so each sum should come in 3 possibilities and hence three 0's and three 6's.

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9
of 13 vote

what about writing, 12-17, so that possibility of coming 1-12 will be equal (always zero)

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1
of 1 vote

``that's great... :-)  if anyone will ask me this question, i'll say this first.. lolz``

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0

+1, thats pretty smart and creative :)

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3
of 3 vote

The probability generating function sum(pix^i) of the even distribution {1,2,...,12} is 1/12(x+x^2+...+x^12)=G(x)H(x) where G(x)=1/6(x+x^2+...+x^6), where G(x) is the probability generating function of the regular fair dice.

Solving for H(x), you get H(x)=1/2*(x^0+x^6)=1/6(3x^0+3x^6) so the unmarked dice has 3 0's and 3 6's.

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1
of 1 vote

``````someone please point me to some theory links for these ( probability ) kind of questions.
specifically for interview's perspective``````

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0
of 0 vote

Agreed, 3 0s and 3 6s is the correct answer.

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0
of 0 vote

What about the numbering be 6,0,12,13,14,15 . In this case every number from 1-12 has a probability of 1/36 and thus evenly distributed between 1-12.
Dice 1 Dice 2
6 6
5 0
4 12
3 13
2 14
1 15

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-2
of 2 vote

if including 1 and 12 then three faces 0 and other three 6.
if 1 and 12 is not included then three faces 1 and other three 5.

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0

actually, i'm not sure if your latter solution would work because
num - probabilty
#2 - 3/36
#3 - 3/36
#4 - 3/36
#5 - 3/36
#6 - 6/36
#7 - 6/36
#8 - 3/36
#9 - 3/36
#10 - 3/36
#11 - 3/36

six - regular dice rolls 5 + fixed dice rolls 1 (3/36) AND
regular dice rolls 1+ fixed dice rolls 5 (3/36)
seven- regular dice rolls 6 + fixed dice rolls 1 (3/36) AND
regular dice rolls 2 + fixed dice rolls 5 (3/36)

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