CAT4MBA

Question:

Q2. What is the probability that the sum of the visible dots is highest?

Solution:

We know that in a standard die, the sum of the opposite faces is always equal to 7 and the sum of all the numbers on all the faces = 21

So, when two - dice are stacked on top of the other, the maximum sum that can be obtained = 21 - 1 - 6 - 1 = 13 and the minimum sum that can be obtained = 21 - 1 - 6 - 6 = 8.

Now, using the same logic as above, we can check that the possible sums of the visible dots are: 8, 9, 10, 11, 12 and 13.

Hence the probability of obtaining the highest sum is 1/6.

Thank You.

Ravi Raja

Ravi

We know that in a standard die, the sum of the opposite faces is always equal to 7 and the sum of all the numbers on all the faces = 21

is fine but

So, when two - dice are stacked on top of the other, the maximum sum that can be obtained = 21 - 1 - 6 - 1 = 13 and the minimum sum that can be obtained = 21 - 1 - 6 - 6 = 8.

is not correct I think it should be 42 - 1 - 6 - 1 = 34 (max.) and 42 - 1 - 6 - 6 = 29 (min.)

I am Extremely Sorry for that blunder that I have made.

Thank You so much Anita for pointing out my mistake.

Ravi Raja

. . . . and then the asnwer for the probability question shd also change. . .

let me see !

good that ravi pointed that sum of a pirs of oposite sides is 7.
when two dices are stacked,
for the bottom dice number of visible opposite pairs = 2
so their sum is 7*2 = 14

same for the top dice

so total sum = 28.

so 28 would be constaant througout

the max and minimum would be decided by the top face of the top dice.

so max = 28+6 = 34
min = 28+1 = 29

The probability of getting max = probablilty of getting the top face as 6 for the top dice = 1/ 6