Probability Problem

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yuliawati: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue May 04, 2010 12:36 am; Thanked: 4 times

by yuliawati » Wed Oct 27, 2010 10:26 pm

How to solve this problem quickly? I need more than 5 mins to get the answer. A long calculation....

Source: MGMAT

Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane's score is higher than Bill's, she wins the game. What is the probability the Jane will win the game?

(A) 15/36
(B) 175/432
(C) 575/1296
(D) 583/1296
(E) 1/2

OA later

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Source: Beat The GMAT — Problem Solving | Wed Oct 27, 2010 10:26 pm

Rezinka: Master | Next Rank: 500 Posts; Posts: 131; Joined: Fri Oct 15, 2010 2:39 am; Location: New Delhi, India; Thanked: 11 times

by Rezinka » Wed Oct 27, 2010 11:09 pm

Is OA E?

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yuliawati: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue May 04, 2010 12:36 am; Thanked: 4 times

by yuliawati » Thu Oct 28, 2010 2:42 am

No, the OA isn't E.

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Thu Oct 28, 2010 3:27 am

yuliawati wrote:How to solve this problem quickly? I need more than 5 mins to get the answer. A long calculation....
Source: MGMAT

Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane's score is higher than Bill's, she wins the game. What is the probability the Jane will win the game?

(A) 15/36
(B) 175/432
(C) 575/1296
(D) 583/1296
(E) 1/2

OA later

Yes, this problem involves a long calculation to some extent.
But we can reduce the calculations to some extent by noting that Bill and Jane has equal probability to win the game. But that doesn't mean that the probability is 1/2, as there also exists the case of tie.

Thus, the required probability = [1 - Probability of tie]/2

Now the possible scores are,
2 : In 1 way => (1, 1)
3 : In 2 way => (1, 2), (2, 1)
4 : In 3 way => (1, 3), (2, 2), (3, 1)
5 : In 4 way => (1, 4), (2, 3), (3, 2), (4, 1)
6 : In 5 way => (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
7 : In 6 way => (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
8 : In 5 way => (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
9 : In 4 way => (3, 6), (4, 5), (5, 4), (6, 3)
10 : In 3 way => (4, 6), (5, 5), (6, 4)
11 : In 2 way => (5, 6), (6, 5)
12 : In 1 way => (6, 6)

Now, Probability of tie
= Probability that scores of Bill and Jane is same
= P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12)
= (1 + 4 + 9 + 16 + 25 + 36 + 25 + 16 + 9 + 4 + 1)/(36*36)
= 146/1296

Thus, required probability = [1 - (146/1296)]/2 = 575/1296

The correct answer is C.

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Oct 28, 2010 3:36 am

yuliawati wrote:How to solve this problem quickly? I need more than 5 mins to get the answer. A long calculation....
Source: MGMAT

Bill and Jane play a simple game involving two fair dice, each of which has six sides numbered from 1 to 6 (with an equal chance of landing on any side). Bill rolls the dice and his score is the total of the two dice. Jane then rolls the dice and her score is the total of her two dice. If Jane's score is higher than Bill's, she wins the game. What is the probability the Jane will win the game?

(A) 15/36
(B) 175/432
(C) 575/1296
(D) 583/1296
(E) 1/2

OA later

3 possible outcomes:
P(J) = Probability that Jane wins
P(B) = Probability that Bill wins
P(D) = Probability in a draw (each gets the same score)

P(J) + P(B) + P(D) = 1.
Since Jane and Bill each have the same chance of winning, P(J) = P(B).
Thus:
P(J) + P(J) + P(D) = 1
2*P(J) + P(D) = 1
P(J) = [1- P(D)] / 2

So to determine P(J), we need to determine P(D).

The probabilities of getting the different sums display the following symmetry:

P(2) = P(12)
P(3) = P(11)
P(4) = P(10)
P(5) = P(9)
P(6) = P(8)

Let's determine the various probabilities to discern the pattern:
P(2) = P(1,1) = 1/6 * 1/6 = 1/36.
P(3) = P(1,2) + P(2,1) = 1/36 + 1/36 = 2/36
P(4) = P(1,3) + P(3,1) + P(2,2) = 3/36
Now we see the pattern:
P(5) = 4/36
P(6) = 5/36
P(7) = 6/36

A draw means that each gets the same score:
P(both score 2) = P(2) * P(2) = 1/36 * 1/36 = 1/1296
P(both score 3) = P(3) * P(3) = 2/36 * 2/36 = 4/1296
P(both score 4) = 3/6 * 3/36 = 9/1296
P(both score 5) = 4/36 * 4/36 = 16/1296
P(both score 6) = 5/36 * 5/36 = 25/36

P(both score 2 through 6) = (1+4+9+16+25)/1296 = 55/1296
Thus, P(both score 8 through 12) = 55/1295
P(both score 7) = 6/36 * 6/36 = 36/1296
Thus, P(D) = (55+55+36)/1296 = 146/1296

Thus P(J) = (1 - 146/1296)/2 = 575/1296.

The correct answer is C.

Last edited by GMATGuruNY on Sun Feb 26, 2012 5:08 am, edited 2 times in total.

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Rezinka: Master | Next Rank: 500 Posts; Posts: 131; Joined: Fri Oct 15, 2010 2:39 am; Location: New Delhi, India; Thanked: 11 times

by Rezinka » Thu Oct 28, 2010 3:37 am

Great..
Thanks

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yuliawati: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Tue May 04, 2010 12:36 am; Thanked: 4 times

by yuliawati » Thu Oct 28, 2010 3:49 am

Thank you, Rahul and Mitch!
In dealing with the probability problem, my big mistake is careless calculation. That's very easy to make a mistake^^

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