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30 Oct 2014, 09:52
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
27% (02:35) correct
73%
(02:04)
wrong
based on 490
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 15/36
(B) 175/432
(C) 575/1296
(D) 583/1296
(E) 1/2
08 Oct 2015, 22:53
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Thoughtosphere
There are some calculations involved here but you can make them a bit more efficient:
There are 11 possible sums and 7 is the middle sum. The probability of getting each sum equidistant from 7 will be the same.
Probability of getting sum of 2 (= 1/36) will be the same as the probability of getting sum of 12 (= 1/36) because of the symmetry of outcomes.
Probability of getting a sum of 3 = Probability of getting a sum of 11 = 2/36
and so on...
Probability that both players get a sum of 2 = \((1/36)*(1/36) = 1^2/36^2\)
Probability that both players get a sum of 3 = \((2/36)*(2/36) = 2^2/36^2\)
and so on ...
Note that the sum of squares of first n consecutive positive integers is given by n(n+1)(2n+1)/6
\(1^2/36^2 + 2^2/36^2 + ... + 5^2/36^2 = (1^2 + 2^2 + 3^2 + ..5^2)/36^2 = (5*6*11/6)/36^2 = 55/36^2\)
\(6^2/36^2 = 36/36^2\)
\(5^2/36^2 + 4^2/36^2 + ....1^2/36^2 = 55/36^2\) (from above)
\(Total = (55 + 36 + 55)/36^2 = 146/36^2\)
This is the probability of draw.
30 Oct 2014, 20:58
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Bunuel
I can't believe that this one is a 600 Level Question. It took me around 4 minutes to solve it.
Here's my take:
There are three possible outcomes in the game:
1. B Wins
2. J Wins
3. Draw
So, \(1 = P(J) + P(B) + P(Draw)\)
The probability of B winning or J winning the game is same.
so, \(1 = 2*P(J) + P(Draw)\)
So, we need to calculate the probability of a draw.
A draw can happen when both will have the same sum.
The possible sums are - 2,3,4,5,6,7,8,9,10,11,12
Prob. of each sum is -
2 - 1,1 - 1/36
3 - 1 2, 2 1 - 2/36
4 - 1 3, 2 2, 3 1 - 3/36
5 - 1 4, 2 3, 3 2, 41 - 4/36
6 - 1 5, 2 4, 3 3, 4 2, 5 1 - 5/36
7 - 1 6, 2 5, 3 4, 4 3, 5 2, 61 - 6/36
8 - 2 6, 3 5, 4 4, 5 3, 6 2, - 5/36
9 - 3 6, 4 5, 5 4, 63 - 4/36
10 - 4 6, 5 5, 6 4 - 3/36
11 - 5 6, 6 5 - 2/36
12 - 6 6 - 1/36
Now, since both have to get the same sum, the probability will be same for both and the combined probability will be the square of the probabilities
i.e. 2 - (1/36)^2
3 - (2/36)^2
etc.
Squaring the probabilities and adding them will give 146/1296
So, P(D) = 146/1296
1 = 2*P(J) + P(D)
2*P(J) = 1150/1296
P(J) = 575 / 1296
Ans. C
