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

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CEO
Joined: 21 Jan 2007
Posts: 2736

Kudos [?]: 1023 [0], given: 4

Location: New York City
A gambler has a 1 in 5 chance of winning his first bet, and [#permalink]

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23 Nov 2007, 14:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

Kudos [?]: 1023 [0], given: 4

Senior Manager
Joined: 06 Aug 2007
Posts: 360

Kudos [?]: 34 [0], given: 0

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23 Nov 2007, 15:01
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

Kudos [?]: 34 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3585

Kudos [?]: 4479 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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23 Nov 2007, 15:51
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

Kudos [?]: 4479 [0], given: 360

SVP
Joined: 28 Dec 2005
Posts: 1549

Kudos [?]: 173 [0], given: 2

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23 Nov 2007, 20:16
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

Kudos [?]: 173 [0], given: 2

CEO
Joined: 21 Jan 2007
Posts: 2736

Kudos [?]: 1023 [0], given: 4

Location: New York City

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29 Nov 2007, 13:57
WL or LW

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

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

Kudos [?]: 1023 [0], given: 4

CEO
Joined: 17 Nov 2007
Posts: 3585

Kudos [?]: 4479 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

29 Nov 2007, 14:15
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...

Kudos [?]: 4479 [0], given: 360

CEO
Joined: 21 Jan 2007
Posts: 2736

Kudos [?]: 1023 [0], given: 4

Location: New York City

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29 Nov 2007, 14:22
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.

Kudos [?]: 1023 [0], given: 4

29 Nov 2007, 14:22
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