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

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

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New post 29 Mar 2017, 06:52
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:)
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Mary and Joe are to throw three dice each. The score is the [#permalink]

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New post 29 Mar 2017, 12:49
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



Easiest way to look at this is to find the symmetry in the problem:
Observe that P(3)=P(18), P(4)=P(17),..., P(10)=P(11). In general, P(x)=P(18+3-x) for any x from 3 to 18. (Call it 'Result-1')
Hence,
[1] Prob. of getting 'sum<=10' is Pr1 = P(3)+P(4)+...+P(10)
[2] Prob. of getting 'sum>10' is Pr2 = P(18)+P(17)+..+P(11)
By Result 1, Pr1 = Pr2.
Now, Pr1+Pr2 = 1, which means Pr1=Pr2= 1/2 [Answer is B]

Explanation of Result 1:
D1 = # on dice1
D2 = # on dice2
D3 = # on dice3

sum = D1+D2+D3 = k, where D1, D2, D3 are in [1,6]
=> D1+D2+D3 = 3 + (k-3)
=> (D1-1) + (D2-1) + (D3-1) = (k-3) [call (D1-1) as d1 in next step]
=> d1+d2+d3 = (k-3) where d1,d2,d3 are in [0,5]...... Condition-1

Now, on a different trial let us assume we have a sum = (21-k), where 'k' is same as before & D1, D2, D3 are in [1,6]
D1+D2+D3 = (21-k)
=> D1+D2+D3 = 18-(k-3)
=> (6-D1)+(6-D2)+(6-D3) = (k-3) [call (6-D1) as a1 in next step]
=> a1+a2+a3 = (k-3) where a1,a2,a3 are in [0,5]....... Condition-2

Note that Condition-1 and 2 are same, which means both have exactly same number of solutions in terms of (d1,d2,d3) or (a1,a2,a3).
Now that we have established that # of ways of getting sum 'k' = # of ways of getting sum '21-k', hence the Result-1 mentioned above.
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

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New post 20 Apr 2017, 22:12
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


Hi! Could anybody please help me to find where I am wrong?

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.
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]

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New post 10 Feb 2018, 04:57
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

Answer = 108/216 = 1/2.

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

Thank you!
Re: Mary and Joe are to throw three dice each. The score is the   [#permalink] 10 Feb 2018, 04:57

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