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# Mary and Joe are to throw three dice each. The score is the

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Intern
Joined: 26 Oct 2016
Posts: 32
Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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29 Mar 2017, 07:52
How you approached it?

Like as explained:
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 17 so the probability to get the sum more then 17 (18) is 1/18 (how ? I got stuck in last step though I understood the 1/2 part in original question)[/quote]

Getting 18 = getting 6 on each of the three die = 1/6*1/6*1/6 = 1/216[/quote]

Thanks:)
Intern
Joined: 10 Sep 2016
Posts: 1
Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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20 Apr 2017, 23:12
noboru 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

We can get scores, that are larger than 10 in the following combinations:

11: 452, 443, 335, 551, 641
12: 336, 444, 543, 651, 552
13: 445, 553, 643, 652, 661
14: 554,644,662
15: 654, 555, 663
16: 655, 664
17: 665
18: 666

From those we have 7 combination that can be different in 6 ways, 15 in 3 ways and 3 in 1 way. Multiply each and then add together: 6*7=42, 3*15=45, 3. 45+42+3=90.
Probability: 90/216.
Intern
Joined: 05 Mar 2017
Posts: 2
GMAT 1: 680 Q48 V34
Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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10 Feb 2018, 05:57
Hi, I was wondering if my approach was right.

To get up to 10 score
Possible values 1st dice = 6 (1,2,3,4,5,6)
Possible values 2nd dice = 3 (1,2,3)
Possible values 3rd dice = 1 (1)
Multiply by 3! because the values could be the result of any dice.

Possible ways to get 10 score = 6x3x1x3! = 108
Possible combinations with 3 dices = 6x6x6 = 216

In case the logic is correct, how could I use this approach to calculate other scores (ex. 3, 5 or 8)?

Thank you!
Manager
Joined: 02 Jul 2017
Posts: 56
Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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08 Jun 2018, 23:33

i solved this by using combinations ..
We can total outcomes =6*6*6 =216
I calculated no of cases where sum of dices will be less than or equal to 10 ....i.e for having 10 we can have different arrangements of 1,3,6 & to have sum less than or equal to 10 ..we can have different arrangements of (1),(1,2,3)& (1,2,3,4,5,6) =3!*1way*3ways*6ways=108 ways ..hence P = 1-108/216=1/2 ..

Thanks
Re: Mary and Joe are to throw three dice each. The score is the &nbs [#permalink] 08 Jun 2018, 23:33

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