The faces of each of 7 standard dice are labeled with the integers

Question

The faces of each of 7 standard dice are labeled with the integers from 1 to 6. Let p be the probability that when all 7 dice are rolled, the sum of the numbers on the top faces is 10. What other sum occurs with the same probability as p?

(A) 13
(B) 26
(C) 32
(D) 39
(E) 42

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Nidzo · Accepted Answer

In this instance, the probability will be:  (   The Number of ways in which one can get a value   /   The total possible outcomes   )

As one is rolling 7 standard dice, one knows that:

1)  the smallest sum of the faces that can be rolled is 7 (seven 1's)
 2)  the greatest sum of the faces that can be rolled is 42 (seven 6's)

Looking at this, we know that there is only one single combination of dice which lead to those values.

Next consider the number of ways in which one can get the second smallest and second biggest values:

1)  the second smallest sum of the faces will be 8 (six 1's and a single 2)
 2)  the second greatest sum of the faces will be 41 (six 6's and a single 5)

There are 7 possible ways in which one can have six of one value and one of another value:  \frac{7!}{6!1!}=7

One can see that this starts to become a pattern which continues right through, and if one were to plot this on a graph (x-axis: sum of faces, y-axis: possible ways in getting the sum) one will find that it creates a symmetrical bell curve. This now becomes not a question on probability, rather a question on symmetry.

In other words, each sum of the faces has a corresponding value which is equidistant from the smallest and largest value respectively.

10 is three away from 7, and therefore its corresponding value will be 3 away from 42: 39.

Answer D

Krunaal · Answer

Starting from minimum sum possible,

To get 7 we need (1,1,1,1,1,1,1) => only one way possible
To get 8 we need (1,1,1,1,1,1,2) => 7 ways possible (2 can come on any of the 7 dice)
To get 9 we need (1,1,1,1,1,1,3) OR (1,1,1,1,1,2,2) => 7 + 21 ways possible (7!/6! OR 7!/5!2!)
 To get 10 we need (1,1,1,1,1,1,4) OR (1,1,1,1,1,2,3) => 7 + 42 ways possible (7!/6! OR 7!/5!)

Now, let's start from maximum possible,

To get 42 we need (6,6,6,6,6,6,6) => only one way possible
To get 41 we need (6,6,6,6,6,6,5) => 7 ways possible (5 can come on any of the 7 dice)
To get 40 we need (6,6,6,6,6,6,4) OR (6,6,6,6,6,5,5) => 7 + 21 ways possible (7!/6! OR 7!/5!2!)
 To get 39 we need (6,6,6,6,6,6,3) OR (6,6,6,6,6,4,5) => 7 + 42 ways possible (7!/6! OR 7!/5!)

Testing from both extremes we can see a symmetry, 10 is 3 units far from minimum 7 and 39 is 3 units far from maximum 42, and both have similar possibilities for the sum to occur.

Therefore,  answer D) 39

hharshh · Answer

Minimum value considering all 1 is ( 1x 7=7 )
Now we have to distribute 3 among any dices to have total sum of faces as 10

Similarly, the max value is 7x6 =42 
To have same probability we need 3 to distribute among any 7 dices. Since the sum cannot exceed 42, implies the possible to reduce 3 among 7 dices will give same probability as increasing 3 among 7 dices

Posted from my mobile device

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The faces of each of 7 standard dice are labeled with the integers

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