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# M14 #29 - soccer game probability

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M14 #29 - soccer game probability [#permalink]

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03 Aug 2008, 09:56
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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)

(A) $$\frac{1}{4}$$
(B) $$\frac{3}{10}$$
(C) $$\frac{2}{5}$$
(D) $$\frac{5}{12}$$
(E) $$\frac{1}{2}$$

[Reveal] Spoiler: OA
C

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03 Aug 2008, 10:13
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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
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03 Aug 2008, 10:25
i understand and agree your solution. but, what is wrong with the following logic?

2 teams, either of them can score first.
therefore probability is 1/2.

why is knowing the final score changing the probability?
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03 Aug 2008, 10:37
sset009 wrote:
i understand and agree your solution. but, what is wrong with the following logic?

2 teams, either of them can score first.
therefore probability is 1/2.

why is knowing the final score changing the probability?

two things,
we know that team A scored 3 goals and team B only scored 2 goals...
and all scoring scnarios are equally probable...

consider this, each goal is an event and team A and team B is getting some points on each event
final sore is team A : 3 points and team B : 2 points
So from each goal team A is getting 3/5 points and team B is getting 2/5 points...

If we take equal probability of scoring first goal by each team... we are have to give a high chance to team A to score rest of the goals to maintain 3:2 score, --> team A has to score more points in later goals to reach at 3 points...

i hope i dint confuse you more
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03 Aug 2008, 17:59
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

Don D, i vaguely remember the formula that if there are n items to be arranged of which x are of type 1 and y are of type 2, then the number of ways it can be done is n!/(x!*y!) . Is this the right formula ?
Could you ps also show the rationale for this formula ? Also does this hold for only two types or multiple types also ?
eg. n!/(x!*y!*z!)
One more thing.. can you also tell me the expression and logic for circular permutations.. how many ways can n different items be arranged in a circle ?
Thanks..
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03 Aug 2008, 18:44
bhushangiri wrote:
Don D, i vaguely remember the formula that if there are n items to be arranged of which x are of type 1 and y are of type 2, then the number of ways it can be done is n!/(x!*y!) . Is this the right formula ?
Could you ps also show the rationale for this formula ? Also does this hold for only two types or multiple types also ?
eg. n!/(x!*y!*z!)
One more thing.. can you also tell me the expression and logic for circular permutations.. how many ways can n different items be arranged in a circle ?
Thanks..

Go through this
http://www.mansw.nsw.edu.au/members/ref ... no4yen.htm
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Re: M14 #29 - soccer game probability [#permalink]

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13 Aug 2009, 04:15
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A simpler way to visualize this problem is as follows.

Assume you are trying to fill 5 slots with 5 marbles or whatever.
There are 3 marbles of type A and 2 of type B.
Now the question boils down to "What is the probability of type B occupying the 1st slot?"
Total number of possibilities for the 1st slot = 5
Desired possibilites (type B) = 2

Probability = Desired/Total = 2/5

Hope that's meaningful.
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Re: M14 #29 - soccer game probability [#permalink]

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05 Oct 2009, 13:34
Just an observation.

Given that the score was 3:2, saying "that all scoring scenarios are equally probable" is the same as saying that the winning team had higher chances of scoring.

That's how the answer can be 2/5.

Otherwise one would assume that each team is equally good and the answer would be 1/2.
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15 Jun 2010, 06:14
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sset009 wrote:
i understand and agree your solution. but, what is wrong with the following logic?

2 teams, either of them can score first.
therefore probability is 1/2.

why is knowing the final score changing the probability?

Saying that the probability is 1/2 is as good as saying you always have 50% of chance of winning a lottery (i.e. You either win or loose). Team with more goals should have more probability of scoring first goal. Consider team A scored 1 million goals (that would be interesting ) and team B did just 1.. who might have scored first?
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Re: M14 #29 - soccer game probability [#permalink]

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16 Jun 2010, 01:08
I think answer should be 1/2

It dosent matter who scored more and who scored less (who win, who lose)
As long as both team have scored atleast 1 goal, probability of scoring first goal by either of the team will always be 1/2

Correct me if i am wrong
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Re: M14 #29 - soccer game probability [#permalink]

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16 Jun 2010, 02:11
hardnstrong wrote:
I think answer should be 1/2

It dosent matter who scored more and who scored less (who win, who lose)
As long as both team have scored atleast 1 goal, probability of scoring first goal by either of the team will always be 1/2

Correct me if i am wrong

That hypothesis would have been true if both the teams are equally skilled.. equally talented.. and equal in everything. But if that would have been the case both the teams would have scored no goals at all.
The fact that they are scoring different goals is an indication of one team is better than the other. Using the very basic definition of probability.. all the events should be equally likely (i.e. unbiased).
To put this definition in perspective.. lets say team A is the best team in the world and all the players of Team B.. have been tied to goal post with a rope. Would you still think they both have same probability of scoring first goal?
In our question, things are not that extreme but we know that one team is better than the other (as it scored more goals) and hence more likely to score the first goal.. If that would not have been the case Michael Jordan and I would have same probability of basketting the ball from other end of the of the court
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Re: M14 #29 - soccer game probability [#permalink]

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16 Jun 2010, 22:56
ialsowant800 wrote:
hardnstrong wrote:
I think answer should be 1/2

It dosent matter who scored more and who scored less (who win, who lose)
As long as both team have scored atleast 1 goal, probability of scoring first goal by either of the team will always be 1/2

Correct me if i am wrong

That hypothesis would have been true if both the teams are equally skilled.. equally talented.. and equal in everything. But if that would have been the case both the teams would have scored no goals at all.
The fact that they are scoring different goals is an indication of one team is better than the other. Using the very basic definition of probability.. all the events should be equally likely (i.e. unbiased).
To put this definition in perspective.. lets say team A is the best team in the world and all the players of Team B.. have been tied to goal post with a rope. Would you still think they both have same probability of scoring first goal?
In our question, things are not that extreme but we know that one team is better than the other (as it scored more goals) and hence more likely to score the first goal.. If that would not have been the case Michael Jordan and I would have same probability of basketting the ball from other end of the of the court

Yes ofcourse you and michael jordon have same probability of basketing the ball (good to boost your confidence ) because either michael will basket it or not (probability 1/2) or you basket or not (probability 1/2)
Its same as flipping the coin, getting head or tail have same probability of 1/2. No matter how skilled is one person
Say you are the world champion in flipping the coin and in a competition you got five heads in a row but still your probability of getting head sixth time will be 1/2. As said by many mathematicians, skills of a person or team have nothing to do with probability. Probability is something is getting the desired outcome divided by total no. of outcomes.

Any say on this
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Re: M14 #29 - soccer game probability [#permalink]

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17 Jun 2010, 02:18
2
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hardnstrong wrote:
ialsowant800 wrote:
hardnstrong wrote:
I think answer should be 1/2

It dosent matter who scored more and who scored less (who win, who lose)
As long as both team have scored atleast 1 goal, probability of scoring first goal by either of the team will always be 1/2

Correct me if i am wrong

That hypothesis would have been true if both the teams are equally skilled.. equally talented.. and equal in everything. But if that would have been the case both the teams would have scored no goals at all.
The fact that they are scoring different goals is an indication of one team is better than the other. Using the very basic definition of probability.. all the events should be equally likely (i.e. unbiased).
To put this definition in perspective.. lets say team A is the best team in the world and all the players of Team B.. have been tied to goal post with a rope. Would you still think they both have same probability of scoring first goal?
In our question, things are not that extreme but we know that one team is better than the other (as it scored more goals) and hence more likely to score the first goal.. If that would not have been the case Michael Jordan and I would have same probability of basketting the ball from other end of the of the court

Yes ofcourse you and michael jordon have same probability of basketing the ball (good to boost your confidence ) because either michael will basket it or not (probability 1/2) or you basket or not (probability 1/2)
Its same as flipping the coin, getting head or tail have same probability of 1/2. No matter how skilled is one person
Say you are the world champion in flipping the coin and in a competition you got five heads in a row but still your probability of getting head sixth time will be 1/2. As said by many mathematicians, skills of a person or team have nothing to do with probability. Probability is something is getting the desired outcome divided by total no. of outcomes.

Any say on this

ok so first to answer the coin flipping competition.. If skills doesn't matter.. what about me having skill to replace the competition coin - without anyone noticing - with my own coin which has 'head' on both side.. would probability of getting a 'tail' will still be 1/2?.. Did you just say that is not fair? well that is exactly my point..

Also to give you another perspective.. what is the probability of getting a six when you throw the fair dice with six faces? Did you say 1/6? Why.. if we apply your logic then either 6 will come or it will not.. and hence probability should be 1/2?

I would recommend that you read the book Probability for Dummies.. to brush-up your concepts. That is really nice book.. attached is snapshot of a relevant page..
Attachments

Probability Misconception.jpg [ 99.45 KiB | Viewed 10949 times ]

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Re: M14 #29 - soccer game probability [#permalink]

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18 Jun 2010, 00:04
It makes sense now .............. thanks for the info
Didnt know the difference between free throw or scoring a goal, and doing thing with fixed number of outcomes (like flipping coin or throwing dice)
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Re: M14 #29 - soccer game probability [#permalink]

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20 Jun 2011, 09:26
THERE ARE 2 C AND 3 T
NUMBER OF OPTINS:
5!/3!*2!

T CAN BE FIRST IN 4 OPTIONS:
TCCTT
TTCCT
TTTCC
TCTCT
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Re: M14 #29 - soccer game probability [#permalink]

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21 Jun 2012, 05:57
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It's a number's game. To fill the first slot, the losing team has a 2 in 5 chances of having scored the first goal, since there are 5 goals in total and 2 goals were scored by the loser.

Alternatively, the total arrangements of goals can be seen as WWW (3 goals for the winning team) and LL (2 goals for losing team). The total number of ways of arranging WWWLL, taking account of identical items is 5!/ (3!*2!)=10 ways.

Since the first spot goes to the losing team, there will be a total of 4 goals to be arranged (WWWL) and the number of ways is 4!/(3!)= 4 ways.

Therefore probability of losing team scoring first is # favorable outcomes/# total possible outcomes = 4/10 = 2/5.

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Re: M14 #29 - soccer game probability [#permalink]

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21 Jun 2012, 22:24
Hi All

I choose 1/2 .. here probability is for loser team. why the score of winning team need to be considered so it will be 1 out of 2 goals ..

Can u plz explain me

Thank u
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Re: M14 #29 - soccer game probability [#permalink]

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22 Jun 2012, 01:42
MBACHANGE wrote:
Hi All

I choose 1/2 .. here probability is for loser team. why the score of winning team need to be considered so it will be 1 out of 2 goals ..

Can u plz explain me

Thank u

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)

A. $$\frac{1}{4}$$
B. $$\frac{3}{10}$$
C. $$\frac{2}{5}$$
D. $$\frac{5}{12}$$
E. $$\frac{1}{2}$$

Consider empty slots for 5 goals: *****. Say W is a goal scored by the winner and L is a goal scored by the loser. We need the probability that when distributing these goals (5 letters LLWWW) into 5 slots L comes first.

Since there are 2 L's out of total 5 letters then P=Favorable/Total=2/5.

Hope it's clear.
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Re: M14 #29 - soccer game probability [#permalink]

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22 Jun 2012, 02:10
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
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Re: M14 #29 - soccer game probability [#permalink]

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22 Jun 2012, 02:16
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Expert's post
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.
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Re: M14 #29 - soccer game probability   [#permalink] 22 Jun 2012, 02:16

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# M14 #29 - soccer game probability

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