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M14 #29 - soccer game probability

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Intern
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Re: M14 #29 - soccer game probability [#permalink] New post 21 Jun 2013, 06:52
Winning Team scored 3/5, winning team did not score first >> 1 - 3/5 = 2/5?
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Re: M14 #29 - soccer game probability [#permalink] New post 13 Feb 2014, 06:33
Bunuel wrote:
MBACHANGE wrote:
Thanks bunel

what if question was loser team scored second goal instead of first goal. then also it will 2/5 rgt? and if winning team scored first, second or third goal will be 3/5

Regards

Regards


Absolutely, the probability that the loser team scored the first, second, ..., fifth goal is the same and equals to 2/5 and the probability that the winner team scored the first, second, ..., fifth goal is the same and equals to 3/5.

Check similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-105990.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html

Hope it helps.


Guys the only difference here between 2/5=4/10 and 1/2 is that since we already know the score the probabilities change. If we had to decide at the beginning of the match then yes probability is 1/2 but after with new info it is 2/5

Hope this clarifies
Cheers
J
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Re: soccer game probability [#permalink] New post 10 Jun 2014, 10:48
durgesh79 wrote:
sset009 wrote:
If a certain soccer game ended 3:2, what is the probability that the side that lost scored first?

(Assume that all scoring scenarios are equiprobable)

(C) 2008 GMAT Club - m14#29

* \frac{1}{4}
* \frac{3}{10}
* \frac{2}{5}
* \frac{5}{12}
* \frac{1}{2}


2 teams A and B ... A won and B lost..

the problems is similar where we have 5 slots, A can take 3 and B can take 2 postions

all possible combinations = 5!/(2! * 3!) = 10

fav cambinations we know that first slot is B.. so rest slots are 4 and A can take 3 postions and B can take 1
= 4! / (3! * 1!) = 4

Probability = 4/10 = 2/5 option C


But when you say that fav cambinations we know that first slot is B.. so rest slots are 4 and A can take 3 postions and B can take 1
= 4! / (3! * 1!) = 4

arent you assuming that the two goals scored by b are the same?

if they are not same then it should be 2 * 4! / (3!*1!) right?
Re: soccer game probability   [#permalink] 10 Jun 2014, 10:48
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