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# 5 women are in a race. What is the probability that

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CEO
Joined: 21 Jan 2007
Posts: 2734

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Location: New York City
5 women are in a race. What is the probability that [#permalink]

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19 Nov 2007, 00:09
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Question Stats:

100% (00:49) correct 0% (00:00) wrong based on 9 sessions

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5 women are in a race.

What is the probability that Sally and Jenny both will not win the race?

What is the probability that Sally, but not Jenny, will win the race?

Kudos [?]: 1076 [0], given: 4

CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4672 [0], given: 360

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

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19 Nov 2007, 00:18
all cases: 5!

1. x ? ? ? ? (x is one of 3)

P=n/N= 4!*3/5!=3/5

2. 1 ? ? ? ?

p=m/N=4!/5!=1/5

Kudos [?]: 4672 [0], given: 360

Director
Joined: 26 Feb 2006
Posts: 900

Kudos [?]: 161 [0], given: 0

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19 Nov 2007, 00:48
walker wrote:
all cases: 5!

1. x ? ? ? ? (x is one of 3)

P=n/N= 4!*3/5!=3/5

2. 1 ? ? ? ?

p=m/N=4!/5!=1/5

confusing and not clear but i concur with you.

Kudos [?]: 161 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4672 [0], given: 360

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

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19 Nov 2007, 01:03
I'll try to explain

all cases: 5! - total number of all combinations.

1. x ? ? ? ? (x is one of 3)

1st,2nd,3th,4th,5th
x - is not Sally and Jenny. 5-2=3 - another women
? - any woman beside x.
? ? ? ? - n=4! combinations.
therefore,

P=n/N= 4!*3/5!=3/5

2. 1 ? ? ? ?

1 - is Sally.
? - any woman beside 1.
? ? ? ? - n=4! combinations.
therefore,

p=m/N=4!/5!=1/5
[/i]

looks better

Kudos [?]: 4672 [0], given: 360

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1076 [0], given: 4

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22 Nov 2007, 19:36
1.

Prob that Sally will not win = 4/5
Prob that Jeny will not win = 4/5

Hence 4/5 * 4/5 = 16/25

correct?

Kudos [?]: 1076 [0], given: 4

Senior Manager
Joined: 06 Aug 2007
Posts: 360

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Re: combinatorics - Sally and Jenny [#permalink]

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22 Nov 2007, 20:44
bmwhype2 wrote:
5 women are in a race.

What is the probability that Sally and Jenny both will not win the race?

What is the probability that Sally, but not Jenny, will win the race?

Here is my explanation:

1. for the first one its simple..what is the probability of both of them winning is 2/5

So the probability of them not winning is 1 - 2/5 = 3/5

2. for the second one...read the question in another way what is the probability of Sally winning

its 1/5 - I dont see why do you have to consider part of Jenny not winning.

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Intern
Joined: 25 Nov 2007
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01 Dec 2007, 13:16
bmwhype2 wrote:
1.

Prob that Sally will not win = 4/5
Prob that Jeny will not win = 4/5

Hence 4/5 * 4/5 = 16/25

correct?

I think this is the short cut way

Prob that Sally will not win = 4/5
Prob that Jeny will not win = 3/4(As Sally is already out of equation)

Hence 4/5 * 3/4 = 3/5

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SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1089 [0], given: 5

Location: New York
Re: combinatorics - Sally and Jenny [#permalink]

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26 Aug 2008, 21:45
bmwhype2 wrote:
5 women are in a race.

What is the probability that Sally and Jenny both will not win the race?

What is the probability that Sally, but not Jenny, will win the race?

1)
p for Sally win the race = 1/5
p for Jenny win the race =1/5

p for all others win the race = 1-(1/5+1/5) =3/5

2) Sally win the race =1/5

What is OA?
_________________

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Kudos [?]: 1089 [0], given: 5

Intern
Joined: 23 Feb 2006
Posts: 5

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27 Aug 2008, 06:38
bmwhype2 wrote:
1.

Prob that Sally will not win = 4/5
Prob that Jeny will not win = 4/5

Hence 4/5 * 4/5 = 16/25

correct?

Can someone explain the flaw here?

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SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1089 [1], given: 5

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27 Aug 2008, 07:11
1
KUDOS
manOnFire wrote:
bmwhype2 wrote:
1.

Prob that Sally will not win = 4/5
Prob that Jeny will not win = 4/5

Hence 4/5 * 4/5 = 16/25

correct?

Can someone explain the flaw here?

5 women are in a race.
What is the probability that Sally and Jenny both will not win the race?

Say S,J, X1,X2,X5 are in the race.. [ assuming the only one person will win the race out of 5 with equal probablity]
All these events[Winning S, Winning J, Winning X1.] are "mutually exclusive" --> Out comes are not common.
P(S) = 1/5
P(J) =1/5
P(X1)=1/5
P(X2)=1/5
P(X3)=1/5

probability of either one of them winning hte race
= P(S or J or X1 or X2 or X3) =P(S) +P(J)+P(X1)+P(X2)+P(X3) =1
probability that Sally and Jenny both will not win the race = probability that either one of X1 or X2 or X2 them win the race = P(X1)+P(X2)+P(X3) =1/5+1/5+1/5=3/5

check the below link, it may be useful for you.
http://www.regentsprep.org/regents/math/mutual/Lmutual.htm
_________________

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Senior Manager
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Re: combinatorics - Sally and Jenny [#permalink]

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16 Feb 2010, 13:41
bmwhype2 wrote:
5 women are in a race.

What is the probability that Sally and Jenny both will not win the race?

What is the probability that Sally, but not Jenny, will win the race?

Guess the question wording is not correct logically...

2 people can not win a race together... hence instead of saying..."probability that Sally and Jenny both will not win the race".. it should be "probability that Sally or Jenny both will not win the race"

Even with the second question - it should be - "What is the probability that Sally, will win the race?"... we don't need to specify not Jenny.. as Sally and Jenny cannot win the race at the same time
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Kudos [?]: 419 [0], given: 47

Re: combinatorics - Sally and Jenny   [#permalink] 16 Feb 2010, 13:41
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# 5 women are in a race. What is the probability that

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