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SVP  Joined: 21 Jan 2007
Posts: 1856
Location: New York City
In a fantasy football game ended 3:2, what is the  [#permalink]

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2
6 00:00

Difficulty:   65% (hard)

Question Stats: 58% (01:54) correct 42% (01:14) wrong based on 148 sessions

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In a fantasy football game ended 3:2, what is the probability that the side that lost scored first?

A. 1/4
B. 3/10
C. 4/10
D. 5/12
E. 1/2

OPEN DISCUSSION OF THIS QUESTION IS HERE: m14-29-soccer-game-probability-68335.html
Manager  Joined: 02 Aug 2007
Posts: 101

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First get the total number of possible arrangements of 5 goals.
Permutations forumula with repeating elements
Total = 5!/3!*2! = 10 possible arrangements

Losing team in 1 position
Possible arrangments
Losing = 4!/3!1! = 4

so 4/10 C.
Director  Joined: 28 Mar 2006
Posts: 877
Re: Challenge - Ratios and Probability  [#permalink]

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bmwhype2 wrote:
In a fantasy football game ended 3:2, what is the probability that the side that lost scored first?

1/4
3/10
4/10
5/12
1/2

Let winning team goals be W,W,W and losing teams be L, L

So we need total # os combinations with 1 L at the 1st poisition

For eg., LWWWL

So total # of possibilities = 5!/3!*2! = 10

Now coming to the first position L we cannot distinguish

so having 1 L in 1st position can be done in 4!/3! (since there are 3 Ws) = 4

So probability = 4/10
CEO  Joined: 07 Jul 2004
Posts: 2890
Location: Singapore

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Call the two teams A and B. A won 3:2.

We want
BBAAA -> We need to arrange just the last 4 letters keeping B in the first position.

# of ways to do this is 4!/3! = 4
Total # of ways to arrange this word = 5!/2!3! = 10

P = 4/10 = 2/5
VP  Joined: 07 Nov 2007
Posts: 1060
Location: New York
Re: Challenge - Ratios and Probability  [#permalink]

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bmwhype2 wrote:
In a fantasy football game ended 3:2, what is the probability that the side that lost scored first?

1/4
3/10
4/10
5/12
1/2

WWLLL
-- total no of ways = 5!/2!3!

[W]WLLL -- first WIN is fixed..
no of ways = 4!/3!

P= 4/10
Manager  Joined: 22 Jul 2008
Posts: 90
Re: Challenge - Ratios and Probability  [#permalink]

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What is the difference between this question and the following question?

In 5 flips of a coin, there are are 2 Heads and 3 Tails. What is the prob. that a Head appeared first?
Just as each flip of a coin is an independent event, so is the scoring of each goal.
Total no. of possible events is 2^5 in the case of a coin flip. Shouldn't the same be the case with each goal scored? Of course we know that one team scored 3 and the other, 2 goals. But, either of the 2 teams could have scored each goal. Therefore, total possible outcomes should be 2^5 = 32. Now, if L appears 1st, the total probable outcomes can be 1*(4 C 1) = 4.

Hence Prob. should be = 4/2^5 = 1/8.
What's wrong with the answer or the explanation?
Of course, the other explanations also sound right and the answer does appear to be 4/10.

P.S. Mind you! 5!/2!*3! is not the total possible outcomes but is the total probable outcomes, i.e., there are 10 ways in which 3Ws and 2Ls can be arranged. And 4 of those 10 ways are the ways in which 1 of the 2 Ls can appear in the 1st position.
VP  Joined: 07 Nov 2007
Posts: 1060
Location: New York
Re: Challenge - Ratios and Probability  [#permalink]

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KASSALMD wrote:
What is the difference between this question and the following question?

In 5 flips of a coin, there are are 2 Heads and 3 Tails. What is the prob. that a Head appeared first?
Just as each flip of a coin is an independent event, so is the scoring of each goal.
Total no. of possible events is 2^5 in the case of a coin flip. Shouldn't the same be the case with each goal scored? Of course we know that one team scored 3 and the other, 2 goals. But, either of the 2 teams could have scored each goal. Therefore, total possible outcomes should be 2^5 = 32. Now, if L appears 1st, the total probable outcomes can be 1*(4 C 1) = 4.

Hence Prob. should be = 4/2^5 = 1/8.
What's wrong with the answer or the explanation?
Of course, the other explanations also sound right and the answer does appear to be 4/10.

P.S. Mind you! 5!/2!*3! is not the total possible outcomes but is the total probable outcomes, i.e., there are 10 ways in which 3Ws and 2Ls can be arranged. And 4 of those 10 ways are the ways in which 1 of the 2 Ls can appear in the 1st position.

In 5 flips of a coin, there are are 2 Heads and 3 Tails. What is the prob. that a Head appeared first?
It should be the same as foot ball..
why are you getting 2^5 out comes..

You want probablity for "head appeared first" out of "2 Heads and 3 Tails occurred"
NOT all possible out comes.

HHTTT --> can be arranged in 5!/2!3!
[H]HTTT -- can be arrange din 4!/3!

p is same ..
Manager  Joined: 15 Jul 2008
Posts: 147
Re: Challenge - Ratios and Probability  [#permalink]

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Okay here is where you are confusing the two...

KASSALMD wrote:
What is the difference between this question and the following question?

In 5 flips of a coin, there are are 2 Heads and 3 Tails. What is the prob. that a Head appeared first?
Just as each flip of a coin is an independent event, so is the scoring of each goal.
Total no. of possible events is 2^5 in the case of a coin flip. No. This would be true only if nothing has happened yet. In the question you already know that there are 3 heads and two tails. So the total number of possible ways you could have got this 3/2 split is 5!/(3!2!)

Shouldn't the same be the case with each goal scored? Of course we know that one team scored 3 and the other, 2 goals. But, either of the 2 teams could have scored each goal. Therefore, total possible outcomes should be 2^5 = 32. Now, if L appears 1st, the total probable outcomes can be 1*(4 C 1) = 4.
Again that would be true if nothing has happened yet and you compute the probability of first even starting with tail and later getting a 3/2 split. But we already know 3/2 split and the question is asking for probability of one particular order of event from the total possible order of events for 3/2 split. This can be explained by the concepts of priori and posterior. But i do not remember that properly and hence wont go that way.
Hence Prob. should be = 4/2^5 = 1/8.
What's wrong with the answer or the explanation?
Of course, the other explanations also sound right and the answer does appear to be 4/10.
Manager  Joined: 22 Jul 2008
Posts: 90
Re: Challenge - Ratios and Probability  [#permalink]

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Never mind! My explanation is valid for the question: What is the probability of getting 2 Heads in a toss of 5 coins with a Head being the 1st to appear? This is different from: what is the prob. of getting a Head 1st if one gets 3 Tails and 2 Heads in a flip of 5 coins?
Manager  Joined: 15 Jul 2008
Posts: 147
Re: Challenge - Ratios and Probability  [#permalink]

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KASSALMD wrote:
Never mind! My explanation is valid for the question: What is the probability of getting 2 Heads in a toss of 5 coins with a Head being the 1st to appear? This is different from: what is the prob. of getting a Head 1st if one gets 3 Tails and 2 Heads in a flip of 5 coins?

Exactly what we tried to explain..
Intern  Joined: 30 Jun 2009
Posts: 33
Re: Challenge - Ratios and Probability  [#permalink]

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Guys, I am not very good at those "#of ways possible" questions.
Could you please explain when do we have to use the 5! for example and why do we have to divide by 3!

Sorry for that question that might sound stupid...

Thx
Intern  Joined: 23 Aug 2009
Posts: 27
Re: Challenge - Ratios and Probability  [#permalink]

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1
defoue wrote:
Guys, I am not very good at those "#of ways possible" questions.
Could you please explain when do we have to use the 5! for example and why do we have to divide by 3!

Sorry for that question that might sound stupid...

Thx

defoue

There is a formula for the no of permutations of n different things, out of which Q1 are alike and are of one type, Q2 are alike and are of 2nd type, Q3 are alike and are of 3rd type, and rest all different:
No of permutations = n!/(Q1!*Q2!*Q3!)

for eg: no of words formed with defoue is:
6!/2! (as e is 2 times)

Hope it is clear to you now.
Intern  Joined: 30 Jun 2009
Posts: 33
Re: Challenge - Ratios and Probability  [#permalink]

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saurabhricha wrote:
defoue wrote:
Guys, I am not very good at those "#of ways possible" questions.
Could you please explain when do we have to use the 5! for example and why do we have to divide by 3!

Sorry for that question that might sound stupid...

Thx

defoue

There is a formula for the no of permutations of n different things, out of which Q1 are alike and are of one type, Q2 are alike and are of 2nd type, Q3 are alike and are of 3rd type, and rest all different:
No of permutations = n!/(Q1!*Q2!*Q3!)

for eg: no of words formed with defoue is:
6!/2! (as e is 2 times)

Hope it is clear to you now.

Very clear!!!
Could you please explain what is difference between combination and permutation.
Intern  Joined: 23 Aug 2009
Posts: 27
Re: Challenge - Ratios and Probability  [#permalink]

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defoue wrote:
saurabhricha wrote:
defoue wrote:
Guys, I am not very good at those "#of ways possible" questions.
Could you please explain when do we have to use the 5! for example and why do we have to divide by 3!

Sorry for that question that might sound stupid...

Thx

defoue

There is a formula for the no of permutations of n different things, out of which Q1 are alike and are of one type, Q2 are alike and are of 2nd type, Q3 are alike and are of 3rd type, and rest all different:
No of permutations = n!/(Q1!*Q2!*Q3!)

for eg: no of words formed with defoue is:
6!/2! (as e is 2 times)

Hope it is clear to you now.

Very clear!!!
Could you please explain what is difference between combination and permutation.

In simple words,

Combinations: The number of ways in which r things can be SELECTED out of n different things. In Combinations, ORDER is not important.
Foe e.g: Selection of 2 people from 10 different people.

Permutations: The number of ways in which r things can be SELECTED & ARRANGED out of n different things. Here, ORDER is important.
For e.g: Selection of 2 people from 10 people for the post of Manager and Director. Here order is important. (Just an example)
Manager  Joined: 27 Oct 2008
Posts: 125
Re: Challenge - Ratios and Probability  [#permalink]

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In a fantasy football game ended 3:2, what is the probability that the side that lost scored first?

1/4
3/10
4/10
5/12
1/2

Soln:
Let the teams be A and B.
for a goal ratio of 3: 2 we can imagine it to be following arrangement. AAABB
where A represents that team A has scored and B represents that team B has scored.

So the total number of arrangements is = 5!/(3! * 2!) = 10 ways

Now to find the number of ways where the team B scores first, we can fix B to first position.
Thus we have BAAAB, and of this the first B alone is fixed. So the others can be arranged in
= 4!/3! ways
= 4

Thus the probability is = 4/10
Intern  Joined: 26 Aug 2010
Posts: 17
Re: Challenge - Ratios and Probability  [#permalink]

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3
We have 5 goals in the match. The question is to define the probability that the first goal was scored by the second team, which scored 2 goals in total. Thus, P=2/5 or 4/10.
I think there is no reason for complex calculations.
Manager  Joined: 16 Jun 2010
Posts: 118
Re: Challenge - Ratios and Probability  [#permalink]

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Hey Kronax .. pretty cool.
Manager  Joined: 02 Apr 2012
Posts: 58
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Re: Challenge - Ratios and Probability  [#permalink]

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srivas wrote:
In a fantasy football game ended 3:2, what is the probability that the side that lost scored first?

1/4
3/10
4/10
5/12
1/2

Soln:
Let the teams be A and B.
for a goal ratio of 3: 2 we can imagine it to be following arrangement. AAABB
where A represents that team A has scored and B represents that team B has scored.

So the total number of arrangements is = 5!/(3! * 2!) = 10 ways

Now to find the number of ways where the team B scores first, we can fix B to first position.
Thus we have BAAAB, and of this the first B alone is fixed. So the others can be arranged in
= 4!/3! ways
= 4

Thus the probability is = 4/10

Ok.
With the same method, the probability that the team B (3 points) scored first is:
We have A BAAB, and of this the first A alone is fixed. So the others can be arranged in
= $$\frac{4!}{(2!*2!)}$$ ways

= $$\frac{(4*3*2)}{(2*2)}$$

= 6

Result:$$\frac{6}{10}$$

Probabilities that A or B scored first are:

$$\frac{4}{10} + \frac{6}{10} = \frac{10}{10}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 65764
Re: Challenge - Ratios and Probability  [#permalink]

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1
Maxirosario2012 wrote:
srivas wrote:
In a fantasy football game ended 3:2, what is the probability that the side that lost scored first?

1/4
3/10
4/10
5/12
1/2

Soln:
Let the teams be A and B.
for a goal ratio of 3: 2 we can imagine it to be following arrangement. AAABB
where A represents that team A has scored and B represents that team B has scored.

So the total number of arrangements is = 5!/(3! * 2!) = 10 ways

Now to find the number of ways where the team B scores first, we can fix B to first position.
Thus we have BAAAB, and of this the first B alone is fixed. So the others can be arranged in
= 4!/3! ways
= 4

Thus the probability is = 4/10

Ok.
With the same method, the probability that the team B (3 points) scored first is:
We have A BAAB, and of this the first A alone is fixed. So the others can be arranged in
= $$\frac{4!}{(2!*2!)}$$ ways

= $$\frac{(4*3*2)}{(2*2)}$$

= 6

Result:$$\frac{6}{10}$$

Probabilities that A or B scored first are:

$$\frac{4}{10} + \frac{6}{10} = \frac{10}{10}$$

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: m14-29-soccer-game-probability-68335.html
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Re: In a fantasy football game ended 3:2, what is the  [#permalink]

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# In a fantasy football game ended 3:2, what is the Question banks Downloads My Bookmarks Reviews Important topics  