Baseball's World Series matches 2 teams against each other in a best-of-seven 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%
Pls explain.
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In a best-of-seven series, at most 7 games will be played. Thus:beat_gmat_09 wrote:Baseball's World Series matches 2 teams against each other in a best-of-seven 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%
Pls explain.
P(fewer than 7 games) + P(exactly 7 games) = 1
So P(fewer than 7 games) = 1 - P(exactly 7 games).
To determine P(exactly 7 games), let's say that the teams playing are team A and team B. The problem implies that each team has an equal chance of winning a game. The only way that 7 games will be played is if A wins 3 of the first 6 games and B wins the other 3.
P(AAABBB) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/64
Since any arrangement of AAABBB will yield a good outcome, we have to account for all the different ways we could arrange AAABBB: 6!/(3!*3!) = 20
Multiplying the results above, we see that:
P(each team wins 3 of the first 6 games) = 1/64 * 20 = 5/16.
Thus, P(exactly 7 games) = 5/16.
So P(fewer than 7 games) = 1 - 5/16 = 11/16 = 68.75%.
The correct answer is D.
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Great explanation, simple and easy to understand !GMATGuruNY wrote:In a best-of-seven series, at most 7 games will be played. Thus:beat_gmat_09 wrote:Baseball's World Series matches 2 teams against each other in a best-of-seven 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%
Pls explain.
P(fewer than 7 games) + P(exactly 7 games) = 1
So P(fewer than 7 games) = 1 - P(exactly 7 games).
To determine P(exactly 7 games), let's say that the teams playing are team A and team B. The problem implies that each team has an equal chance of winning a game. The only way that 7 games will be played is if A wins 3 of the first 6 games and B wins the other 3.
P(AAABBB) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/64
The number of ways to arrange AAABBB = 6!(3!*3!) = 20
Multiplying, we see that:
P(each team wins 3 of the first 6 games) = 1/64 * 20 = 5/16.
Thus, P(exactly 7 games) = 5/16.
So P(fewer than 7 games) = 1 - 5/16 = 11/16 = 68.75%.
The correct answer is D.
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- goyalsau
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Can you please explain it a bit further,GMATGuruNY wrote: In a best-of-seven series, at most 7 games will be played. Thus:
P(fewer than 7 games) + P(exactly 7 games) = 1
So P(fewer than 7 games) = 1 - P(exactly 7 games).
1 = all the cases that are possible, ,
IF we try to do it like this There are two means A and B,
LIKE A team won first 4 maches, ( ONE CASE )
A team won First 4 matches out of 5 matches, { !5/ !4 = 5 ways, }
A team won first 4 matches out of 6 matches { !6/!4 * !2 = 15 ways }
Now i am not getting the answer,
I m not to understand that this particular line
P(fewer than 7 games) + P(exactly 7 games) = 1
Saurabh Goyal
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P(event happens) + P(event doesn't happen) = 1.goyalsau wrote:Can you please explain it a bit further,GMATGuruNY wrote: In a best-of-seven series, at most 7 games will be played. Thus:
P(fewer than 7 games) + P(exactly 7 games) = 1
So P(fewer than 7 games) = 1 - P(exactly 7 games).
1 = all the cases that are possible, ,
IF we try to do it like this There are two means A and B,
LIKE A team won first 4 maches, ( ONE CASE )
A team won First 4 matches out of 5 matches, { !5/ !4 = 5 ways, }
A team won first 4 matches out of 6 matches { !6/!4 * !2 = 15 ways }
Now i am not getting the answer,
I m not to understand that this particular line
P(fewer than 7 games) + P(exactly 7 games) = 1
In a best-of-seven series, either 4, 5, 6, or 7 games are played.
P(event happens) = P(fewer than 7 games) = P(4, 5 or 6 games)
P(event doesn't happen) = P(not fewer than 7 games) = P(exactly 7 games)
Since either fewer than 7 games are played (meaning 4, 5, or 6 games) or exactly 7 games are played, the sum of the two probabilities above must be 1:
P(fewer than 7 games) + P(exactly 7 games) = 1
P(fewer than 7 games) involves quite a bit of arithmetic:
P(4 games) means that A or B won all of the first 4 games.
P(5 games) means that A or B won 4 out of the first 5 games, but not the first 4, because then only 4 games would be played.
P(6 games) means that A or B won 4 out of the first 6 games, but not the first 4, because then only 4 games would be played, and not 4 out of the first 5, because then only 5 games would be played.
It's much quicker -- and much easier -- to determine P(exactly 7 games) and subtract this fraction from 1.
Clear?
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- goyalsau
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P(event happens) + P(event doesn't happen) = 1.
Thanks Guru i got it,
But Don't know where to use it next time, or not to use it,
Thanks Guru i got it,
But Don't know where to use it next time, or not to use it,
Saurabh Goyal
[email protected]
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Hi Guru, everything makes perfect sense in your explanation, but I don't understand one thing. What happened to the 7th game? You never accounted for that. Why is that?
Thanks!
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You actually do account for the seventh game, since your basic equation isIM_onthegrind wrote:Hi Guru, everything makes perfect sense in your explanation, but I don't understand one thing. What happened to the 7th game? You never accounted for that. Why is that?
Thanks!
P(<7 games) = 1 - P(=7 games)
So it's best to compute P(=7 games) -- the probability of exactly 7 games -- then subtract that from 1 to find the answer you're after.
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We are given that two teams are to play in a best-of-seven series and the probability of either team winning a game is 1/2. We need to determine the probability that the world series will have fewer than 7 games.beat_gmat_09 wrote:Baseball's World Series matches 2 teams against each other in a best-of-seven 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%
We can start by using the following formula:
1 = P(the world series will have less than 7 games) + P(the world series will have 7 games)
Thus,
P(the world series will have less than 7 games) = 1 - P(the world series will have 7 games)
Let's determine the probability that the world series will have 7 games.
We can denote each winning team as team "A" and team "B". There is only one option to get to 7 games: team A wins three games and team B wins three games ( forcing a game 7).
P(A-A-A-B-B-B) = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64
However, we need to determine in how many ways team A can win three games and team B can win three games. That number will be equivalent to how many ways we can organize the letters A-A-A-B-B-B.
We use the indistinguishable permutations formula to determine the number of ways to arrange A-A-A-B-B-B: 6!/(3! x 3!) = 20 ways
Each of these 20 ways has the same probability of occurring. Thus, the total probability is:
20(1/64) = 20/64 = 5/16
The probability of the world series having 7 games is 5/16. Thus the probability of the world series having less than 7 games is 1 - 5/16 = 11/16 = 68.75%
Answer: D
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