noboru wrote:

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

A. 24/64

B. 32/64

C. 36/64

D. 40/64

E. 42/64

Easiest way to look at this is to find the symmetry in the problem:

Observe that P(3)=P(18), P(4)=P(17),..., P(10)=P(11). In general, P(x)=P(18+3-x) for any x from 3 to 18. (Call it 'Result-1')

Hence,

[1] Prob. of getting 'sum<=10' is Pr1 = P(3)+P(4)+...+P(10)

[2] Prob. of getting 'sum>10' is Pr2 = P(18)+P(17)+..+P(11)

By Result 1, Pr1 = Pr2.

Now, Pr1+Pr2 = 1, which means Pr1=Pr2= 1/2 [Answer is B]

Explanation of Result 1:

D1 = # on dice1

D2 = # on dice2

D3 = # on dice3

sum = D1+D2+D3 = k, where D1, D2, D3 are in [1,6]

=> D1+D2+D3 = 3 + (k-3)

=> (D1-1) + (D2-1) + (D3-1) = (k-3) [call (D1-1) as d1 in next step]

=> d1+d2+d3 = (k-3) where d1,d2,d3 are in [0,5]......

Condition-1Now, on a different trial let us assume we have a sum = (21-k), where 'k' is same as before & D1, D2, D3 are in [1,6]

D1+D2+D3 = (21-k)

=> D1+D2+D3 = 18-(k-3)

=> (6-D1)+(6-D2)+(6-D3) = (k-3) [call (6-D1) as a1 in next step]

=> a1+a2+a3 = (k-3) where a1,a2,a3 are in [0,5].......

Condition-2Note that Condition-1 and 2 are same, which means both have exactly same number of solutions in terms of (d1,d2,d3) or (a1,a2,a3).

Now that we have established that # of ways of getting sum 'k' = # of ways of getting sum '21-k', hence the Result-1 mentioned above.