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kevincan
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The Red Jackets and Blue Blazers are playing for the 09 Mar 2007, 08:04
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The Red Jackets and Blue Blazers are playing for the championship. The first team to win a total of four matches is declared the winner and no more matches are played. If the two teams are equally likely to win any match and no match ends in a draw, what is the probability that at least six matches will be played?

(A) 3/8 (B) 7/16 (C) 1/2 (D) 9/16 (E) 5/8

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The Red Jackets and Blue Blazers are playing for the 09 Mar 2007, 08:59
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Hi,
The prob that each of the teams will win first 4 games is (1/2)^4 or 1/16
So the prob that the game will end in 4 games is (1/16)+(/16)1/8
The prob that the game will end in 5 games is 4/16 for team A and 4/16 for team B and the required prob is 4/16+4/16=1/2
The prob that the winner will be selected in 4 or 5 games is 1/8+1/2 or 5/8
The prob that 6 or more games will be played is 1-(5/8)=3/8

So 3/8 is my choice
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The Red Jackets and Blue Blazers are playing for the 11 Mar 2007, 04:10
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Man, I was trying to solve this prob using combinatorics... :?
like in order for there to be at least six games played, a team has to at least have won 2 games, and expand from there...

BG's answer is pretty much straight forward I think. It seems correct. :-D

Guess I have a long way to go to get 51 in math :cry:
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The Red Jackets and Blue Blazers are playing for the 11 Mar 2007, 04:20
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BG
Hi,
The prob that each of the teams will win first 4 games is (1/2)^4 or 1/16
So the prob that the game will end in 4 games is (1/16)+(/16)1/8
The prob that the game will end in 5 games is 4/16 for team A and 4/16 for team B and the required prob is 4/16+4/16=1/2
The prob that the winner will be selected in 4 or 5 games is 1/8+1/2 or 5/8
The prob that 6 or more games will be played is 1-(5/8)=3/8

So 3/8 is my choice
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How did you get 4/16 as the probability to end the series in 5 games?...
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The Red Jackets and Blue Blazers are playing for the 11 Mar 2007, 04:32
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SimaQ
BG
Hi,
The prob that each of the teams will win first 4 games is (1/2)^4 or 1/16
So the prob that the game will end in 4 games is (1/16)+(/16)1/8
The prob that the game will end in 5 games is 4/16 for team A and 4/16 for team B and the required prob is 4/16+4/16=1/2
The prob that the winner will be selected in 4 or 5 games is 1/8+1/2 or 5/8
The prob that 6 or more games will be played is 1-(5/8)=3/8

So 3/8 is my choice

How did you get 4/16 as the probability to end the series in 5 games?...
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To end the series in 5 games, the winning team has to have won 4 games and lost one. There are 4 ways of winning 4 games and losing one:
l w w w w
w l w w w
w w l w w
w w w l w

(not w w w w l cuz if the team won 4 straight games, it'd end there!)

and the prob of a team winning 4 games is (1/2)^4 = 1/16

Therefore, (1/16)x4 = 4/16
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The Red Jackets and Blue Blazers are playing for the 11 Mar 2007, 05:52
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I am going to give this a try, I get 9/16.

The probability asked is:

1 - (P(one team wins the first 4 games) + P(one team wins 4 games in the first 5 games)

= 1 - 2* (4C4 * (1/2)^4 + 5C4*(1/2)^4*(1/2) ) - multiplying by 2 since it can happen to either team.

= 1 - (2 /16+ 5/16) = 1-7/16 = 9/16.
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The Red Jackets and Blue Blazers are playing for the 11 Mar 2007, 06:03
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Just saw ricokevin's response. Ok, so it looks like when I calculated the probability of winning 4 in 5 games, I have included again, the probability of winning 4 games straight.



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