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M08-21

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Intern
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Re M08-21  [#permalink]

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New post 17 Jan 2019, 08:12
I think this is a high-quality question and I agree with explanation.
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Re: M08-21  [#permalink]

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New post 13 Apr 2019, 06:29
I came up with a generalized sum of dice probability rule based on this problem:

Probability of rolling a sum of x with n # of rolls

P(x) = P(7n - x)

e.g.

3 rolls

P(3) = P(18)


2 rolls

P(3) = P(11)
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M08-21  [#permalink]

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New post 22 May 2019, 08:03
The sum of three dice will always be between - 3*1 and 6*3
Using this favorable outcome are 8 [11,12,13,14...18]
Total outcomes - 16 [3,4,5,6,7....17,18]

Required probability - 8/16=1/2=32/64
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Re: M08-21  [#permalink]

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New post 04 Aug 2019, 00:13
I solved it this way. The range of dice sum is 3-18. Losing combinations is 3-10 (10 inclusive)(range 8) and winning combinations is 11-18(range 8). so the probability will be 50%. as we have equal range on both sides.

please let me know if my logic is correct.

Bunuel 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. \(\frac{24}{64}\)
B. \(\frac{32}{64}\)
C. \(\frac{36}{64}\)
D. \(\frac{40}{64}\)
E. \(\frac{42}{64}\)

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Re: M08-21   [#permalink] 04 Aug 2019, 00:13

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