naveendevaraj 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
Answer: B.
Explanations are most welcomed .Thanks in advance
This is a one liner on CAT Test in India, and the answer is ½ or B. 32/64 as per the choices supplied; but it could be a nightmare for most of the GMAT aspirants who are either missing the phenomenon or the hidden take away as explained under:
In other words, we are required to find the probability that Joe will throw a sum more than 10 when he throws three dice.
Since, there are 3 through 18 i.e. 16 different sums that could be had from the three dice, and those 16 different sums can be had in 6^3 = 216 dissimilar ways.
The engendering function is (x + x^2 + x^3 + x^4 + x^5 + x^6) ^3
= [x (1 - x^6)] ^3/ (1 - x) ^3
= x^3 (1 - 3 x^6 + 3 x^12 - x^18) (1 - x) ^ (-3)
= (x^3 - 3 x^9 + 3 x^15 - x^21) {1 + C (3, 1) x + C (4, 2) x^2 + C (5, 3) x^3 +...}, and we require the coefficients of x^11 through x^18 to add up.
Appearance of x^11: x^3 C (10, 8) x^8 - 3 x^9 C (4, 2) x^2, and the required coefficient is C (10, 8) - 3 C (4, 2) = 45 - 18 = 27.
Appearance of x^12: x^3 C (11, 9) x^9 - 3 x^9 C (5, 3) x^3, and the required coefficient is C (11, 9) - 3 C (5, 3) = 55 - 30 = 25.
Appearance of x^13: x^3 C (12, 10) x^10 - 3 x^9 C (6, 4) x^4, and the required coefficient is C (12, 10) - 3 C (6, 4) = 66 - 45 = 21.
...and so on. We would notice that the 16 different sums would be occurring in the following peculiar pattern of the number of ways (this could be the take away). Read:
Sum -> Number of ways
3 -> 1
4 -> 3
5 -> 6
6 -> 10
7 -> 15
8 -> 21
9 -> 25
10 -> 27
11 -> 27
12 -> 25
13 -> 21
14 -> 15
15 -> 10
16 -> 6
17 -> 3
18 -> 1
So it's the matter of first half and next half only.
[spoiler]
Fifty-fifty praaji fifty-fifty...:D
B[/spoiler]
