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Mary and Joe throw three dice each. If Mary gets 10, what

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Mary and Joe throw three dice each. If Mary gets 10, what [#permalink] New post 04 Apr 2005, 13:08
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Mary and Joe throw three dice each. If Mary gets 10, what is the probability that Joe will outscore Mary?

OPEN DISCUSSION OF THIS QUESTION IS HERE: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html AND HERE: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html
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 [#permalink] New post 05 Apr 2005, 09:02
1/4
my formulas are too lengthy to be those you should actually use to solve this, so I'm waiting for other people to answer. (provided that my answer is correct!)
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 [#permalink] New post 05 Apr 2005, 11:05
that is incorrect, the correct answer is 1/2...any ideas how to get to that?
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 [#permalink] New post 05 Apr 2005, 14:00
rc1979 wrote:
that is incorrect, the correct answer is 1/2...any ideas how to get to that?


damn! this is not testing our knowledge of permutations/combinations/possible outcomes
I couldn't be more stupid
Follow this:
expected value of a dice= 3.5
expected value of 3 ones= 10.5
so we have 1/2 to score less than 10.5 and 1/2 to score more than 10.5
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Re: probability [#permalink] New post 05 Apr 2005, 22:37
gmat2me2 wrote:
rc1979 wrote:
Mary and Joe throw three dice each. If Mary gets 10, what is the probability that Joe will outscore Mary?


How to do this in a concise way?


see above :-D
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Re: probability [#permalink] New post 06 Apr 2005, 03:35
thearch wrote:
gmat2me2 wrote:
rc1979 wrote:
Mary and Joe throw three dice each. If Mary gets 10, what is the probability that Joe will outscore Mary?


How to do this in a concise way?


see above :-D


Thanks but can you explain how you got 3.5 for 1 dice ?
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 [#permalink] New post 06 Apr 2005, 03:52
expected value results from the sum of probability*value (it is an aleatory number)
in this case:
1/6*1+1/6*2+1/6*3+1/6*4+1/6*5+1/6*6=1/6*(1+2+3+4+5+6)=3.5
another example could be
we have 5 bottles of wine that are sold in a supermarket
2 of type A
2 of type B
1 of type C
bottles of type A are charged $5 each
bottles of type B are charged $10 each
bottles of type C are charged $15 each
John buys a bottle. If John doesn't know what bottle he chose, how much should he expect to pay?
EV=2/5*$5+2/5*$10+1/5*$15=$9

don't know if I'm clear with my English. please correct me if my reasoning is wrong :-D
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 [#permalink] New post 06 Apr 2005, 03:53
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Proba is 1/2 just because we are half way from 18.
From 3 (minimum score) to 10 there is a 7 range
from 11 (minimum score to outscore) to 18 (max) you have a 7 range.

So Prob to score 11 = Prob to score 10
Prob 9 = Prob 12
...
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 [#permalink] New post 06 Apr 2005, 04:30
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1/2

min result : 3 (1+1+1)
max result : 18 (6+6+6)
number of outcomes : 18-3+1 = 16

We want outcome >10 and max is 18 so there are 8 possibilities over 16

8/16 = 1/2
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 [#permalink] New post 07 Apr 2005, 09:36
Good Antmavel!

Total possible values are from 3-18, therefore 16 outcomes.
Desired outcomes are from 11-18, therefore 8 outcomes.
Possibility is 8/16=1/2.
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 [#permalink] New post 28 Dec 2007, 07:45
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bmwhype2 wrote:
Antmavel wrote:
1/2

min result : 3 (1+1+1)
max result : 18 (6+6+6)
number of outcomes : 18-3+1 = 16

We want outcome >10 and max is 18 so there are 8 possibilities over 16

8/16 = 1/2


whats the logic behind that equation?


3 dice: lowest number possible 1+1+1 =3, greatest number possible 6+6+6=18

18-3+1 (because inclusive) = 16 possible outcomes. 8 of which are greater than 10
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Re: probability [#permalink] New post 26 Aug 2008, 08:21
There is no need for any calculations here. No matter what the numbers are, the probability will always be 1/2. The reason for that is that Joe can have only 2 outcomes- either outscore Mary or be outscored. Simple, isn't it! It would have been a different question if it was required to find out probability of Joe getting a particular score.
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Re: probability [#permalink] New post 26 Aug 2008, 09:52
KASSALMD wrote:
There is no need for any calculations here. No matter what the numbers are, the probability will always be 1/2. The reason for that is that Joe can have only 2 outcomes- either outscore Mary or be outscored. Simple, isn't it! It would have been a different question if it was required to find out probability of Joe getting a particular score.


In this case calculations are not required. because possible outcomes for sum >10 equal to possible coutcomes for sum <=10

What if question is changed to .
Mary and Joe throw three dice each. If Mary gets 8, what is the probability that Joe will outscore Mary?

Definitely you need to calculate.
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Re: probability [#permalink] New post 06 Jun 2009, 04:08
Guys can u explain the Ist question( Probability )

wat i did is as follows
Total no cases= 6*6*6= 216
Favourable no of cases= any no greater than 10 till 18 that is 11 12 13 14 15 16 17 18= 8

so pro= 8/216= 1/27

is it ok? :|
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Re: probability [#permalink] New post 08 Jun 2009, 02:51
rc1979 wrote:
Mary and Joe throw three dice each. If Mary gets 10, what is the probability that Joe will outscore Mary?


Assuming D6

6*6*6 VALUES in total

What's Pr (sum>10?)
What's 1-Pr(sum<8)?

1<=X<=6
1<=Y<=6
1<=Z<=6

X+Y+Z < 8
|{X,Y,Z}|?

Let's calculate Pr(sum<8):

Say X=1
Then Y+Z<7?
1/5
1/4
1/3
1/2
1/1
2/4
2/3
2/2
2/1
3/3
3/2
3/1
4/2
4/1
5/1
(15 values)

Say X=2
Then Y+Z<6?
4/1
3/2
3/1
2/3
2/2
2/1
1/4
1/3
1/2
1/1
(10 values)

say x=3
Then Y+Z<5?
3/1
2/2
2/1
1/3
1/2
1/1
(6 values)

say x=4
Then Y+Z<4?
2/1
1/2
1/1
(3 values)

say x=5
Then Y+Z<3?
1/1
(1 value)

say x=6?
then Y+Z<2? (0 values)

In total: 15+10+6+3+1+0 = 35/216

1-Pr(sum<8) = 1-35/216 = 181/216

Hence Probability is 181/216 ~= 0.8379
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Re: Re: [#permalink] New post 08 Jun 2009, 02:57
eschn3am wrote:
bmwhype2 wrote:
Antmavel wrote:
1/2

min result : 3 (1+1+1)
max result : 18 (6+6+6)
number of outcomes : 18-3+1 = 16

We want outcome >10 and max is 18 so there are 8 possibilities over 16

8/16 = 1/2


whats the logic behind that equation?


3 dice: lowest number possible 1+1+1 =3, greatest number possible 6+6+6=18

18-3+1 (because inclusive) = 16 possible outcomes. 8 of which are greater than 10



but certain sums come up more frequently than others... ie you get 3 (1+1+1) 1/216 of the time but 4 (1+1+2,1+2+1,2+1+1) 3/216 = 1/72 of the time...
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Re: probability [#permalink] New post 24 Jun 2009, 00:35
KASSALMD wrote:
There is no need for any calculations here. No matter what the numbers are, the probability will always be 1/2. The reason for that is that Joe can have only 2 outcomes- either outscore Mary or be outscored. Simple, isn't it! It would have been a different question if it was required to find out probability of Joe getting a particular score.


This is incorrect. You are right that they are two outcomes "of interest", but they are NOT either outscore Mary or be outscored. They are outscore mary and NOT be outscored (i.e., can tie). And just because there are two outcomes of interest doesn't mean they have equal probability.
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Re: probability [#permalink] New post 24 Jun 2009, 01:27
Hades is true. A perfect answer.
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Re: Mary and Joe throw three dice each. If Mary gets 10, what [#permalink] New post 22 Mar 2013, 06:23
not able to understand this ques and any of the solution mentioned above..
if anyone can explain me in some other way ...then please explain :(
thanks in advance
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Re: Mary and Joe throw three dice each. If Mary gets 10, what [#permalink] New post 22 Mar 2013, 06:26
Expert's post
Perhaps wrote:
not able to understand this ques and any of the solution mentioned above..
if anyone can explain me in some other way ...then please explain :(
thanks in advance


Check here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html or here: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.

Also, please be more specific when asking a question. Thank you.
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Re: Mary and Joe throw three dice each. If Mary gets 10, what   [#permalink] 22 Mar 2013, 06:26
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