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15 Sep 2014, 23:54



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Re: M1429
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14 Aug 2016, 07:14
Hi,
I could not understand why we have only two arrangements? I understand that there are only two Ls and occurring first can happen in two ways but what is conceptually wrong in considering these three options LWWLW, LLWWW & LWLWW? In all previous options loser is scoring first.
Please explain.
Regards Yash



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22 Aug 2016, 03:56
I think this is a highquality question and I agree with explanation.



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23 Sep 2016, 06:32
Yash26 wrote: Hi,
I could not understand why we have only two arrangements? I understand that there are only two Ls and occurring first can happen in two ways but what is conceptually wrong in considering these three options LWWLW, LLWWW & LWLWW? In all previous options loser is scoring first.
Please explain.
Regards Yash You can take it this way: LLWWW can be arranged in 5!/(3!*2!), which is 10 Fixing the first goal to be scored by L, we get 4 slots remaining with LWWW, which can be arranged by 4!/3! leading to 4. So, probability is 4/10 or 2/5.



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15 Dec 2016, 06:10
Bunuel wrote: Official Solution:
If a certain soccer game ended 3:2, what is the probability that the side that lost scored first? (Assume that all scoring scenarios have the same probabiliy)
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 of when \(L\) comes first while distributing these goals (5 letters \(LLWWW\)) into 5 slots. Since there are 2 \(L\)'s out of total 5 letters, then \(P=\frac{Favorable}{Total}=\frac{2}{5}\).
Answer: C Wow! Lesson learned. I was so set on using combinations to solve this that I completely missed the obvious and easy approach. Took me 3 mins to ensure that my probability was correct when I should have setup the answer the way you did and be done within 30 seconds.
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Bunuel 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 have the same probabiliy)
A. \(\frac{1}{4}\) B. \(\frac{3}{10}\) C. \(\frac{2}{5}\) D. \(\frac{5}{12}\) E. \(\frac{1}{2}\) Total goals = 5 Winner goals = 3 Loser goals = 2 P (Loser hit the first goal) = P (Winner did not hit the first goal) = 1  P (Winner hit the first goal) => 1 3/5 => 2/5 C
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Bunuel wrote: Official Solution:
If a certain soccer game ended 3:2, what is the probability that the side that lost scored first? (Assume that all scoring scenarios have the same probabiliy)
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 of when \(L\) comes first while distributing these goals (5 letters \(LLWWW\)) into 5 slots. Since there are 2 \(L\)'s out of total 5 letters, then \(P=\frac{Favorable}{Total}=\frac{2}{5}\).
Answer: C Is the reason why this is not 1/2 because the potential outcomes have already been dictated by the score? In other words, if the question stem had not provided the score of the game and the question was "what is the probability that Team A scores first?", then that probability would be 1/2? However, because you are given the final score, there are 10 possible ways the game occurred: LLWWW LWLWW LWWLW LWWWL WWWLL WWLWL WWLWL WWLLW WLWWL WLLWW WLWLW The first four outcomes are the only ways the team could have not gotten the first goal out of the ten outcomes. P = 4/10 = 2/5 Correct? P.S.  Can someone please express this in a combinatoric approach? I understand that there are 5c2 ways to arrange LLWWW, but how do we get the numerator to be 4? I believe it's (2c1)*2, since order of "which" L doesn't matter, but can someone please confirm?



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07 Feb 2017, 05:40
brooklyndude wrote: Bunuel wrote: Official Solution:
If a certain soccer game ended 3:2, what is the probability that the side that lost scored first? (Assume that all scoring scenarios have the same probabiliy)
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 of when \(L\) comes first while distributing these goals (5 letters \(LLWWW\)) into 5 slots. Since there are 2 \(L\)'s out of total 5 letters, then \(P=\frac{Favorable}{Total}=\frac{2}{5}\).
Answer: C Is the reason why this is not 1/2 because the potential outcomes have already been dictated by the score? In other words, if the question stem had not provided the score of the game and the question was "what is the probability that Team A scores first?", then that probability would be 1/2? However, because you are given the final score, there are 10 possible ways the game occurred: LLWWW LWLWW LWWLW LWWWL WWWLL WWLWL WWLWL WWLLW WLWWL WLLWW WLWLW The first four outcomes are the only ways the team could have not gotten the first goal out of the ten outcomes. P = 4/10 = 2/5 Correct? P.S.  Can someone please express this in a combinatoric approach? I understand that there are 5c2 ways to arrange LLWWW, but how do we get the numerator to be 4? I believe it's (2c1)*2, since order of "which" L doesn't matter, but can someone please confirm? Yes, that's correct.
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Re: M1429
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01 Jul 2017, 02:36
I made a combinatrics approach.
Say 5 Goals, winning team is A and loosing team is B
So total combination of goals is 5!/(3!2!) Since A has three goals and B has 2 Goals
Total is 10.
Question asked what is the probability that B scored first. So,
B, (3 combinations of A and 1 Combinations of B)
(3 combinations of A and 1 Combinations of B) implies, 4 goals of which 3 is A and 1 is B
4!/(3!1!)= 4
Therefore 4/10=2/5! Tada!



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20 Apr 2018, 10:27
Denominator: Number of ways to arrange “LLWWW”: 5! / 2!(3!) = 10 Numerator: Number of ways to arrange “LLWWW” assuming the first “L” is set: 4! / 1!(3!) = 4 Probability: 4/10 = 2/5










