Stumped by a probability problem

[naveendevaraj]

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

[ksundar]

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.

[sanju09]

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]

[sanju09]

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.

[GMATGuruNY]

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).

[Brent@GMATPrepNow]

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

First consider rolling 1 die.
If you were to roll a die millions of times, what would be the average value rolled?
Well, since each outcome (1,2,3,4,5 and 6) are all equally likely, the average of the outcomes will be 3.5 (since (1+2+3+4+5+6)/6 = 3.5)
Of course, it's impossible to roll 3.5, but notice that the 3.5 divides the outcomes into two parts. We have the numbers less than 3.5 (that is 1,2,3) and the numbers greater than 3.5 (that is 4,5,6).

Also notice that, in one roll, P(rolling less than 3.5) = 1/2, and P(rolling more than 3.5) = 1/2

Now consider rolling 3 dice.
If the average expected outcome is 3.5 when one die is rolled, the average expected sum will be 10.5 when three dice are rolled (since 3.5 + 3.5 + 3.5 = 10.5)

IMPORTANT: If 10.5 is the average expected sums, then half of all sum will be less than 10.5 and half will be greater than 10.5. In other words, P(sum is less than 10.5) = 1/2 and P(sum is greater than 10.5) = 1/2

The question asks us to find P(sum is greater than 10). This is the same as P(sum is greater than 10.5), which means this probability = 1/2

Answer: B

Cheers,
Brent