A roll of three dice
Arjuna is going to roll three fair ten-sided dice with faces labelled 0 , 1 , 2 , … , 9 .
First he rolls two dice, and finds the sum of the two rolls.
Then he rolls the third die.
If the probability that the sum of the first two rolls equals the third roll can be expressed as b a , where a , b are coprime positive integers, then what is a + b ?
The answer is 211.
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2 solutions
Assume third dice shows C .First and second dice rolls A and B respectively.
We need to C = A + B . C can vary from 0 → 9 .
For a roll C we have number of solutions equals ( 2 − 1 C + 2 − 1 ) = C + 1
Total required solutions = ∑ C = 0 9 C + 1 = 1 + 2 + 3 + . . . . + 1 0 = 5 5
All possible solutions = 1 × 1 0 × 1 0 = 1 0 0 0
Required probability 1 0 0 0 5 5 = 2 0 0 1 1
The third die has 10 possibilities, each occurring with probability of 0.1.
There is only one way to get 0: 0+0.
Two ways to get 1: 0+1, 1+0.
Three ways to get 2: 0+2, 1+1, 2+0.
This pattern continues until the ten ways to get 9: 0+9, 1+8, 2+7, ..., 9+0
So the number of hits is a sum: 1+2+3+...+10 = 55
Number of possibilities is 1000.
Fraction 1 0 0 0 5 5 = 2 0 0 1 1