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A gambler has a 1 in 5 chance of winning his first bet, and

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CEO
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A gambler has a 1 in 5 chance of winning his first bet, and [#permalink] New post 23 Nov 2007, 13:41
A gambler has a 1 in 5 chance of winning his first bet, and a 1 in 7 chance of winning his second bet, and 1 in 7 chance of winning his third bet. What is his chance of winning exactly one of the bets if they are independent?
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Re: gambling probability [#permalink] New post 23 Nov 2007, 14:01
bmwhype2 wrote:
A gambler has a 1 in 5 chance of winning his first bet, and a 1 in 7 chance of winning his second bet, and 1 in 7 chance of winning his third bet. What is his chance of winning exactly one of the bets if they are independent?


I think there are 2 possible answers:

1. 36/245

2. 24/245
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 [#permalink] New post 23 Nov 2007, 14:51
Expert's post
12/35

p=1/5*6/7*6/7+4/5*1/7*6/7+4/5*6/7*1/7=(6*6+2*4*6)/(5*7*7)=

=6*14/(5*7*7)=12/35
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 [#permalink] New post 23 Nov 2007, 19:16
walker wrote:
12/35

p=1/5*6/7*6/7+4/5*1/7*6/7+4/5*6/7*1/7=(6*6+2*4*6)/(5*7*7)=

=6*14/(5*7*7)=12/35


bingo. i didnt reduce, so i ended up with 84/245
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 [#permalink] New post 29 Nov 2007, 12:57
WL or LW

1/5*6/7 = 6/35
4/5*1/7 = 4/35

6/35 + 4/35= 10/35 = 2/7
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 [#permalink] New post 29 Nov 2007, 13:15
Expert's post
bmwhype2 wrote:
WL or LW

1/5*6/7 = 6/35
4/5*1/7 = 4/35

6/35 + 4/35= 10/35 = 2/7


it is suitable for two bets but we have three ones...
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 [#permalink] New post 29 Nov 2007, 13:22
walker wrote:
bmwhype2 wrote:
WL or LW

1/5*6/7 = 6/35
4/5*1/7 = 4/35

6/35 + 4/35= 10/35 = 2/7


it is suitable for two bets but we have three ones...


sorry. i was looking at a different version of this problem.

should be 84/245, using the same method.
  [#permalink] 29 Nov 2007, 13:22
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