GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 03 Aug 2020, 10:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# In a fantasy football game ended 3:2, what is the

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

### Show Tags

30 Oct 2007, 11:05
2
6
00:00

Difficulty:

65% (hard)

Question Stats:

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

### HideShow timer Statistics

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

### Show Tags

30 Oct 2007, 11:18
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]

### Show Tags

30 Oct 2007, 17:17
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

### Show Tags

31 Oct 2007, 00:58
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]

### Show Tags

26 Aug 2008, 11:44
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]

### Show Tags

27 Aug 2008, 06:50
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]

### Show Tags

27 Aug 2008, 07:12
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]

### Show Tags

27 Aug 2008, 07:15
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]

### Show Tags

27 Aug 2008, 08:16
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]

### Show Tags

27 Aug 2008, 08:20
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]

### Show Tags

03 Sep 2009, 04:32
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]

### Show Tags

03 Sep 2009, 05:02
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]

### Show Tags

03 Sep 2009, 05:31
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]

### Show Tags

03 Sep 2009, 05:44
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]

### Show Tags

28 Sep 2009, 10:20
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]

### Show Tags

27 Sep 2010, 09:20
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]

### Show Tags

27 Sep 2010, 15:12
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]

### Show Tags

18 Jul 2013, 13:37
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
Joined: 02 Sep 2009
Posts: 65764
Re: Challenge - Ratios and Probability  [#permalink]

### Show Tags

18 Jul 2013, 13:48
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
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 15595
Re: In a fantasy football game ended 3:2, what is the  [#permalink]

### Show Tags

02 Jun 2019, 04:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In a fantasy football game ended 3:2, what is the   [#permalink] 02 Jun 2019, 04:47