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Baseball's World Series matches 2 teams against each other
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Baseball's World Series matches 2 teams against each other in a bestofseven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games? (A) 12.5% (B) 25% (C) 31.25% (D) 68.75% (E) 75%
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Originally posted by Titleist on 29 Oct 2003, 10:34.
Last edited by Bunuel on 01 Feb 2014, 04:50, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: Baseball's World Series matches 2 teams against each other
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24 Mar 2014, 02:34
riskietech wrote: Baseball's World Series matches 2 teams against each other in a bestofseven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5% (B) 25% (C) 31.25% (D) 68.75% (E) 75%
I have answer 42.86% but not matching the option.
Explanation: Let us find probability that game will consist of exactly 7 games. So there are 4 wins & 3 losses, but out of these wins, one win has to be last Ex: WWWLLL W possibilities = 6!/(3!*3!) = 20 (dividing so as to eliminate repetitions)
Now for all possible events, there are only four subpossibilities : 1) exactly 4 games played, 2) exactly 5 games played, 3) exactly 6 games played, 4) exactly 7 games played 1) 4 games = total possibilities = 1 (WWWW) 2) 5 games = total possibilities = 4!/(3!*1!) = 4 (for ex: WWWL W) 3) 6 games = total possibilities = 5!/(3!*2!) = 10 (for ex: WWWLL W) 7) 7 games = as explained above = 20 So total possible outcomes = 1 + 4 + 10 + 20 = 35 prob of exact 7 games = 20/35 = 57.14%
prob of fewer than 7 games = 100  57.14 = 42.86% Say the teams are A and B. In order the series to consist of seven games each team has to win 3 games in the first 6 games: {A, A, A, B, B, B}. The probability that A wins is 1/2 and the probability that B wins is also 1/2, so the probability of {A, A, A, B, B, B} is (1/2)^6*6!/(3!3!). Therefore the probability that the series will consist of fewer than 7 games is \(1  (\frac{1}{2})^6*\frac{6!}{3!3!} = 0.6875\). Answer: D.
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The correct answer is 75%.
You have a total of eight different possibilities:
40 in favor of Team A
41 in favor of Team A
42 in favor of Team A
43 in favor of Team A
40 in favor of Team B
41 in favor of Team B
42 in favor of Team B
43 in favor of Team B
So the total number of choices for the game to end with less than 7 games = 6 different possibilities
Therefore % less than 7 games = 6/8 = 75%




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How about this approach.....However I get a different answer
Prob (world series lasts fewer than 7 games) = 1  Prob (world series lasts exactly 7 games)
Prob (Exactly 7 games) => After 6 games, Team A has exactly 3 wins & Team B has exactly 3 wins
Prob (Team A wins a game) = 1/2
Prob (Team B wins a game) = 1/2
Using Binomial probability (Assuming tied games are not possible)
Prob ( Team A or B has exactly 3 wins after 6 games) = 5/16
Prob ( Worls series lasts fewer than 7 games) = 1  5/16 = 11/16 = 68.75 %



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Mathew wrote: How about this approach.....However I get a different answer
Prob (world series lasts fewer than 7 games) = 1  Prob (world series lasts exactly 7 games)
Prob (Exactly 7 games) => After 6 games, Team A has exactly 3 wins & Team B has exactly 3 wins
Prob (Team A wins a game) = 1/2 Prob (Team B wins a game) = 1/2
Using Binomial probability (Assuming tied games are not possible)
Prob ( Team A or B has exactly 3 wins after 6 games) = 5/16
Prob ( Worls series lasts fewer than 7 games) = 1  5/16 = 11/16 = 68.75 %
Hi Mathew,
We can not use binomial probability theorm here. Because in order to use the binomial theorm, the number of trial (or games in this example) should be fixed. In this case, the number of trials (games) are not fixed. It depends on the outcome of each game. For example if team A wins all the first 4 games, then the series ends there..no further games. I think rkrish7's answer 75% is the correct answer.
Thanks



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How about this :
The possibilities are  4,5,6 or 7 games played(who wins is immaterial) =4
Less than 7 games  4,5,6 = 3
Probability =3/4=75%



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I opened a discussion here:
http://www.baseballprimer.com/clutch/ar ... l#comments
Here's what I think is the right answer:
<i>12.5% of the time the series will end in 4 games. (2 in 16)
25% of the time the series will end in 5 games. (8 in 32)
62.5% the series doesn't end in 4 or 5. That means it MUST be 32 after Game 5.
Which means half the time, Game 6 is the last game and it end 42.
Which means that 31.25% of the time, the series will end in 6 games.
12.5% + 25% + 31.25% = 68.75%, what I said earlier.
That leaves 31.25% of the time for it to go 7 games.
Am I being totally obtuse? It's possible. </i>



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In order to determine the probability that the World Series will last less than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.
Let's analyze one way this could happen for Team 1:
Game 1 Game 2 Game 3 Game 4 Game 5 Game 6
T1 Wins T1 Wins T1 Wins T1 Loses T1 Loses T1 Loses
There are many other ways this could happen for Team 1. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for Team 1 to win 3 games and lose 3 games in the first 6 games.
Logically, there are also 20 ways for Team 2 to win 3 games and lose 3 games in the first 6 games.
Thus, there are a total of 40 ways for the World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is.
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:
Thus the probability that the World Series will last less than 7 games is 100%  31.25% = 68.75%.
The correct answer is D.



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Titleist wrote: This was a tough question. Thanks to all of you who put up a great effort!
Dear Sir:
It is not kosher to steal the challenge questions from Manhattan GMAT (which are copyrighted) and post them in a public forum without attribution.
Thanks in advance for your future cooperation.
P.S. Since you have presented this problem as your own, please identify the error in the explanation (the answer is correct, but the explanation is a little faulty).
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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My apologies! Will put down a footnote next time.
btw: the url is: http://www.manhattangmat.com
thanks all!



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Titleist wrote: My apologies! Will put down a footnote next time. btw: the url is: http://www.manhattangmat.comthanks all!
I would prefer that you did not post them at all.
In any case, you are NOT allowed to post the solution as it is private and reserved for the use of our clients. (The solution is password protected and available by subscription only. If you are one of our clients, I don't think Zeke would approve of you sharing his challenge problems and solutions with the world).
BTW, here is the "error" is the solution (the answer is fine, but IMO the explanation is slighty wrong):
"Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for Team 1 to win 3 games and lose 3 games in the first 6 games.
Logically, there are also 20 ways for Team 2 to win 3 games and lose 3 games in the first 6 games. Thus, there are a total of 40 ways for the World Series to last the full 7 games."
While the conclusion that there are 40 ways to go the full seven is correct, it does not follow for the reason given.
The 20 ways for Team 1 to win 3 and lose 3 are the same 20 ways for Team 2 to win 3 and lose 3 (a win for Team 1 is a loss for Team 2). Hence the above method is actually double counting the ways that the two teams can go six games.
There are indeed 40 ways to go SEVEN games, NOT because "there are also 20 ways for Team 2 to win 3 games and lose 3 games...", but because there are two ways to resolve the tie in the seventh game, i.e., Team 1 either wins or loses. Hence, the 20 ways to go six are multipled by the 2 possible outcomes in game 7.
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Hey, AkamaiBrah, You wrote that:
"I would prefer that you did not post them at all.
In any case, you are NOT allowed to post the solution as it is private and reserved for the use of our clients. (The solution is password protected and available by subscription only. If you are one of our clients, I don't think Zeke would approve of you sharing his challenge problems and solutions with the world)."
http://www.manhattangmat.com/ChallProbLastWk.cfm
Well, it might be "private" and "reserved for the use of (your) clients", and the solution might be "password protected" and "available by subscription only", but absolutely anyone can get access to the solution for free, as long as he or she furnishes or fabricates an email address. It's not like Titleist has taken a forpay part of the site and redistributed it.
I am neither a lawyer nor an ethcist, but I fail to see any problem or harm if he takes a free question and reposts it with proper attribution.
If I'm off base, please tell me why.



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AkamaiBrah, I will respect your wishes and not repost any of the "free" questions available at Manhattan GMAT. All I wanted to do was to get other people's input on what the answer was because I was genuinely curious. I had no intention of taking 'credit' for the problem itself as you have accused me of doing. I posted the answer because there are people who participated and were genuinely interested in knowing the 'correct' answer. Again, I do apologize.



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stoolfi wrote: Hey, AkamaiBrah, You wrote that: "I would prefer that you did not post them at all. In any case, you are NOT allowed to post the solution as it is private and reserved for the use of our clients. (The solution is password protected and available by subscription only. If you are one of our clients, I don't think Zeke would approve of you sharing his challenge problems and solutions with the world)." http://www.manhattangmat.com/ChallProbLastWk.cfmWell, it might be "private" and "reserved for the use of (your) clients", and the solution might be "password protected" and "available by subscription only", but absolutely anyone can get access to the solution for free, as long as he or she furnishes or fabricates an email address. It's not like Titleist has taken a forpay part of the site and redistributed it. I am neither a lawyer nor an ethcist, but I fail to see any problem or harm if he takes a free question and reposts it with proper attribution.If I'm off base, please tell me why.
First of all, thank you Titliest for promising to respect the copyright of Manhattan GMAT. I apologize if I came across as a hardass.
Let me address a couple of points:
1) Access to the solutions is not "free" but granted in exchange for your email address. Just because you are not required to pay any money for something, does not make it free. There is an intangible "cost" to you for providing such information and certainly a "value" for MGMAT to obtain it. By publicly posting the solutions, you circumvent the right of MGMAT to receive "payment" for their intellectual property.
2) Copyright protection applies regardless of whether material is "free" or for pay. The network broadcast of a concert is free, but it would be illegal for me to incorporate any significant portion of it in my own work and redistribute it  with or without attribution  without express permission from the copyright holder.
3) While publishing other people's works in part with attribution is ordinarily acceptable (though often abused), publishing other people's works in toto certainly is not. Since MGMAT publishes just one problem per week, I can see a scenario where every problem and solution is published every week, essentially replicating the entire series of problems on a 3rdparty forum. That would not be cool.
4) Just because digital media makes it easy to duplicate and republish material, does not make it right.
While I cannot speak for the owner of MGMAT, IMO an acceptable way to post a challenge question is:
"I found this week's Manhattan GMAT challenge problem interesting and would like to share it with you:"
At which point, the forum is free to discuss the problem and post solutions in their own words. Those who want access to the official solution are welcome to subscribe to the service.
Finally, those of you who have been long time members of this forum may note that some of the challenge problems I post are indeed those from ManhattanGMAT. This is because I compose many of the MGMAT challenge questions and I use this forum as a "test bed" before submitting them to Manhattan GMAT for use on their web site. Hence, you already have VIP access to a significant number of the Manhattan GMAT challenge questions and solutions  long before anyone else in the world. I am willing to continue doing this so long as the users of this forum will respect my copyright, and the copyright of MGMAT once any problem is published on its website.
Thank you.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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1) Access to the solutions is not "free" but granted in exchange for your email address. Just because you are not required to pay any money for something, does not make it free. There is an intangible "cost" to you for providing such information and certainly a "value" for MGMAT to obtain it. By publicly posting the solutions, you circumvent the right of MGMAT to receive "payment" for their intellectual property.
Is someone who furnishes a fake email address committing theft, then?



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AkamaiBrah,
First, I want to say thank you for clarifying your point. I also want to thank you and Manhattan GMAT for coming up with the most challenging questions I have seen from any other test prep service. I appreciate your presence here in this forum as a valueadded resource. I will go ahead and post the weekly challenge problems in the format you have described (if I'm so compelled to do so). What drives me crazy about the M.G. weekly challenge problems is that after I take the usual => 2 mins to solve the problem, I have to wait a week for the answer.
Titleist



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AkamaiBrah wrote: Titleist wrote: My apologies! Will put down a footnote next time. thanks all! I would prefer that you did not post them at all. In any case, you are NOT allowed to post the solution as it is private and reserved for the use of our clients. (The solution is password protected and available by subscription only. If you are one of our clients, I don't think Zeke would approve of you sharing his challenge problems and solutions with the world). BTW, here is the "error" is the solution (the answer is fine, but IMO the explanation is slighty wrong): "Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for Team 1 to win 3 games and lose 3 games in the first 6 games. Logically, there are also 20 ways for Team 2 to win 3 games and lose 3 games in the first 6 games. Thus, there are a total of 40 ways for the World Series to last the full 7 games." While the conclusion that there are 40 ways to go the full seven is correct, it does not follow for the reason given. The 20 ways for Team 1 to win 3 and lose 3 are the same 20 ways for Team 2 to win 3 and lose 3 (a win for Team 1 is a loss for Team 2). Hence the above method is actually double counting the ways that the two teams can go six games. There are indeed 40 ways to go SEVEN games, NOT because "there are also 20 ways for Team 2 to win 3 games and lose 3 games...", but because there are two ways to resolve the tie in the seventh game, i.e., Team 1 either wins or loses. Hence, the 20 ways to go six are multipled by the 2 possible outcomes in game 7. The solution provided is incorrect. This question as written is wrong, and I doubt it is an official GMAT question. A simple binomial probability (.5^7) doesn't work because an event such as LLLLWWW is clearly against the rules. All events must end in a 'W' (in respect to a given team). Since the last component of each event is never in question, it's the same as multiplying by 1. I believe the solution to such a problem should read something like this: (say Cardinals vs. Rangers); There is only one way for the Rangers to sweep the cardinals: 4 consecutive wins (1) There are four ways for the Rangers to win in 5 games (write it out by hand if you like); this is the same as using 4p3 with repeating elements (4!/(3!*1!); (add 4 more) There are 10 ways for the Rangers to win in 6 games; in formula, I believe it would be 5p3 (5!/(3!*2!); (add 10 more) Lastly, there are 20 different ways for the Rangers to win in 7 games; this should be the equivalent of 6p3 (6!/(3!*3!); (add 20 more). Again, the reason why the permutation is one less than the number of games (e.g., permutation of 6 for 7 games played) is because the series must logically end in a W for the winning team. Adding all of these possibilities up, there are 35 ways for the Rangers to win. There are an equal number of ways for the Cardinals to win, so there are 70 feasible ways for the series to play out. So, there are 40 ways that are 7 games played out of 70. 1 (40/70)= 3/7, or ~43%. Using .5^7 to generate 128 possibilities generates 58 nonsensical events (e.g., LLLLLLW).



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Re: Baseball's World Series matches 2 teams against each other
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23 Mar 2014, 12:30
I have answer 42.86% but not matching the option.
Explanation: Let us find probability that game will consist of exactly 7 games. So there are 4 wins & 3 losses, but out of these wins, one win has to be last Ex: WWWLLL W possibilities = 6!/(3!*3!) = 20 (dividing so as to eliminate repetitions)
Now for all possible events, there are only four subpossibilities : 1) exactly 4 games played, 2) exactly 5 games played, 3) exactly 6 games played, 4) exactly 7 games played 1) 4 games = total possibilities = 1 (WWWW) 2) 5 games = total possibilities = 4!/(3!*1!) = 4 (for ex: WWWL W) 3) 6 games = total possibilities = 5!/(3!*2!) = 10 (for ex: WWWLL W) 7) 7 games = as explained above = 20 So total possible outcomes = 1 + 4 + 10 + 20 = 35 prob of exact 7 games = 20/35 = 57.14%
prob of fewer than 7 games = 100  57.14 = 42.86%



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Re: ANSWER
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28 Apr 2014, 10:53
jlgdr wrote: Titleist wrote: In order to determine the probability that the World Series will last less than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.
Let's analyze one way this could happen for Team 1:
Game 1 Game 2 Game 3 Game 4 Game 5 Game 6 T1 Wins T1 Wins T1 Wins T1 Loses T1 Loses T1 Loses
There are many other ways this could happen for Team 1. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for Team 1 to win 3 games and lose 3 games in the first 6 games.
Logically, there are also 20 ways for Team 2 to win 3 games and lose 3 games in the first 6 games.
Thus, there are a total of 40 ways for the World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is.
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:
Thus the probability that the World Series will last less than 7 games is 100%  31.25% = 68.75%.
The correct answer is D. Hi there, The explanation is not complete, could you please post correctly? I didn't quite understand the last part, seems that you are missing something Thanks! Cheers J Here's how I did it, but I don't think its correct So to finish in less than 7 games it will finish in either 4,5 or 6 games. 4 games will be (1/2) ^ 4 = 1/16, one team winning the four games 5 games will be (5C4)(1/2^5) = 5/32 6 games will be (6C4)(1/2^6)=15/64 Adding em up, 29/64 close enough to the correct answer choice Now, i'm not too sure whether I should have considered that for instance in the case of 4 games we would have 2 cases, 1/16 if one team wins and 1/16 if another team wins, so total probability would be 2/16 = 1/8? Same with the other scenarios Could anybody please clarify this? Appreciate it! Thanks! Cheers J



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Re: Baseball's World Series matches 2 teams against each other
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10 May 2014, 05:38
riskietech wrote: Bunuel wrote: riskietech wrote: Baseball's World Series matches 2 teams against each other in a bestofseven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5% (B) 25% (C) 31.25% (D) 68.75% (E) 75%
I have answer 42.86% but not matching the option.
Explanation: Let us find probability that game will consist of exactly 7 games. So there are 4 wins & 3 losses, but out of these wins, one win has to be last Ex: WWWLLL W possibilities = 6!/(3!*3!) = 20 (dividing so as to eliminate repetitions)
Now for all possible events, there are only four subpossibilities : 1) exactly 4 games played, 2) exactly 5 games played, 3) exactly 6 games played, 4) exactly 7 games played 1) 4 games = total possibilities = 1 (WWWW) 2) 5 games = total possibilities = 4!/(3!*1!) = 4 (for ex: WWWL W) 3) 6 games = total possibilities = 5!/(3!*2!) = 10 (for ex: WWWLL W) 7) 7 games = as explained above = 20 So total possible outcomes = 1 + 4 + 10 + 20 = 35 prob of exact 7 games = 20/35 = 57.14%
prob of fewer than 7 games = 100  57.14 = 42.86% Say the teams are A and B. In order the series to consist of seven games each team has to win 3 games in the first 6 games: {A, A, A, B, B, B}. The probability that A wins is 1/2 and the probability that B wins is also 1/2, so the probability of {A, A, A, B, B, B} is (1/2)^6*6!/(3!3!). Therefore the probability that the series will consist of fewer than 7 games is \(1  (\frac{1}{2})^6*\frac{6!}{3!3!} = 0.6875\). Answer: D. I will rephrase your eqn as: prob that exactly 7 games will be played = [6!/(3!*3!)] / [2^6] = 20 / 64 From above what i grasp is total possible outcomes is 2^6 = 64. But this will violate the rule that "The first team to win four games wins the series and no subsequent games are played" I think total possible outcomes(as per the rules) are only 35. Pls correct if I am wrong. We do want to include such cases as WWWWWWW, when we do 1  P(seven games) approach. This case would be one of the cases attributed to WWWW. The probability of this case is (1/2)^4. If you continue this to 7 games you get: WWWW WWW > P=(1/2)^7 WWWW WWL > P=(1/2)^7 WWWW WLW > P=(1/2)^7 WWWW LWW > P=(1/2)^7 WWWW WLL > P=(1/2)^7 WWWW LWL > P=(1/2)^7 WWWW LLW > P=(1/2)^7 WWWW LLL > P=(1/2)^7 Sum = 8*(1/2)^7 = (1/2)^4. The same probability as we got for WWWW. So, 1  P(seven games) would still give the correct answer.
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Re: Baseball's World Series matches 2 teams against each other &nbs
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