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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

Re: Challenge - Ratios and Probability [#permalink]
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. Somebody please resolve the discrepancy.

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.

Re: Challenge - Ratios and Probability [#permalink]
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. Somebody please resolve the discrepancy.

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 .. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: Challenge - Ratios and Probability [#permalink]
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. Somebody please resolve the discrepancy.

Re: Challenge - Ratios and Probability [#permalink]
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?

Re: Challenge - Ratios and Probability [#permalink]
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?

Re: Challenge - Ratios and Probability [#permalink]
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...

Re: Challenge - Ratios and Probability [#permalink]
03 Sep 2009, 05:02

1

This post received KUDOS

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)

Re: Challenge - Ratios and Probability [#permalink]
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.

Re: Challenge - Ratios and Probability [#permalink]
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)

Re: Challenge - Ratios and Probability [#permalink]
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

Re: Challenge - Ratios and Probability [#permalink]
27 Sep 2010, 09:20

3

This post received KUDOS

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.

Re: Challenge - Ratios and Probability [#permalink]
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

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Re: Challenge - Ratios and Probability [#permalink]
18 Jul 2013, 13:48

1

This post received KUDOS

Expert's post

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.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...