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Mary and Joe are to throw three dice each

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Mary and Joe are to throw three dice each [#permalink] New post 06 May 2012, 01:25
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

How to approach this question.
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Re: Mary and Joe are to throw three dice each [#permalink] New post 06 May 2012, 02:04
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subhajeet 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

How to approach this question.


Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.
Expected value of three dice is 3*3.5=10.5.

Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2 = 32/64.

That's because the probability distribution is symmetrical for this case:
The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);
The probability of getting the sum of 4 = the probability of getting the sum of 17;
The probability of getting the sum of 5 = the probability of getting the sum of 16;
...
The probability of getting the sum of 10 = the probability of getting the sum of 11;

Thus the probability of getting the sum from 3 to 10 = the probability of getting the sum from 11 to 18 = 1/2.

Answer: B.

Also discussed here: mother-mary-comes-to-me-86407.html and here: mary-and-joe-126407.html

Hope it's clear.
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Re: Mary and Joe are to throw three dice each [#permalink] New post 06 May 2012, 02:04
P(3) + P(4) + P(5)..........+P(10) = P(11) + P(12) + P(13).......+P(18)

and P(3) + P(4) + ....... + P(18) = 1

Therefore P(11) + P(12) + P(13).......+P(18) = 1/2 or 32/64

Option (B)
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Re: Mary and Joe are to throw three dice each   [#permalink] 06 May 2012, 02:04
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