goyalsau wrote: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).
Can you please explain it a bit further,
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
P(event happens) + P(event doesn't happen) = 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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