Problem Solving

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

There are 216 different possible combinations. Total sum of 3 numbers of each dice can be between 3 and 18. Joe has to get a number greater than 10 which means, out of 16 numbers (sum of 3 dice) possible he can get only 8 numbers greater than 10. 8 is half of 16. So B is the answer.

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]

ksundar wrote:There are 216 different possible combinations. Total sum of 3 numbers of each dice can be between 3 and 18. Joe has to get a number greater than 10 which means, out of 16 numbers (sum of 3 dice) possible he can get only 8 numbers greater than 10. 8 is half of 16. So B is the answer.

This is not always safe to assume.

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

I was asked by PM to offer a "GMAT" solution to this problem.

Let L = lowest possible sum, H = highest possible sum. When a die is rolled multiple times, the probabilities of getting the different sums exhibit the following symmetry:

P(L) = P(H)
P(L+1) = P(H-1)
P(L+2) = P(H-2)
P(L+3) = P(H-3)
etc.

In the problem above, a die is being rolled 3 times, so L=3, and H=18. Plugging into the symmetrical pattern above, we can see that:

P(3) = P(18)
P(4) = P(17)
P(5) = P(16)
P(6) = P(15)
P(7) = P(14)
P(8) = P(13)
P(9) = P(12)
P(10) = P(11)

Thus, we have a 50% chance of getting a sum of 10 or less and a 50% chance of getting a sum greater than 10.

So P(sum greater than 10) = 1/2.

The correct answer is B.

In more general terms:

When a die is rolled multiple times, P(sum below the median sum) = P(sum above the median sum).