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If 2 different representatives are to be selected at random

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Senior Manager
Joined: 02 Dec 2007
Posts: 432
If 2 different representatives are to be selected at random [#permalink]

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05 Sep 2008, 11:03
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If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than1/10 .

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient

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Last edited by Nihit on 05 Sep 2008, 11:15, edited 1 time in total.
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

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05 Sep 2008, 12:06
Nihit wrote:
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than1/10 .

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient

(1) More than 1/2 of the 10 employees are women.
Women count >6
assume 6W 4 M

P = 6C2/10C2 = 60/90=2/3 >1/2
so for W>=6 p>1/2
so for W=9
p= 9C2/10C2= 9*8/10*9 = 72/90 >1/2

not suffcient
(2) The probability that both representatives selected will be men is less than1/10
M 10-M women.

MC2/10C2 <1/10 --> M(M-1)/10*9<1/10
M(M-1)<9 this is possible only when M=4 or less.
So W>=6
same as (1)
not suffcient.

combined same result
E
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Joined: 05 Sep 2008
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05 Sep 2008, 12:41
(1) More than 1/2 of the 10 employees are women.
Women count >6
assume 6W 4 M

P = 6C2/10C2 = 3/5 >1/2
so for W>=6 p>1/2
suffcient
(2) The probability that both representatives selected will be men is less than1/10
M 10-M women.

I DONT FOLLOW YOU HERE, PLEASE FORGIVE ME. ASSUMING 6 OF 10 ARE WOMEN, WOULDNT THAT PROBABILITY BE 6/10*5/9 OR 30/90 WHICH IS 3/9 (SO NOT ABOVE 1/2). ALTERNATIVELY, IT COULD BE ANOTHER EXTREME, 9 OF 10, WHICH WOULD MAKE PROBABILITY 9/10*8/9 WHICH IS 72/90 AND CLEARLY OVER 1/2. IT SEEMS TO ME THAT THIS IS NOT SUFFICIENT. PERHAPS I AM READING QUESTION WRONG? THANKS.

MC2/10C2 <1/10 --> M(M-1)/10<1/10
M(M-1)<1 this is possible only when M=1 or M=0
So W=9 or 10
p= 9C2/10C2 > 1/2
SUFFCIENT
I AGREE WITH YOU HERE, MAKES PERFECT SENSE.
D
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05 Sep 2008, 12:48
grover33 wrote:
(1) More than 1/2 of the 10 employees are women.
Women count >6
assume 6W 4 M

P = 6C2/10C2 = 3/5 >1/2
so for W>=6 p>1/2
suffcient
(2) The probability that both representatives selected will be men is less than1/10
M 10-M women.

I DONT FOLLOW YOU HERE, PLEASE FORGIVE ME. ASSUMING 6 OF 10 ARE WOMEN, WOULDNT THAT PROBABILITY BE 6/10*5/9 OR 30/90 WHICH IS 3/9 (SO NOT ABOVE 1/2). ALTERNATIVELY, IT COULD BE ANOTHER EXTREME, 9 OF 10, WHICH WOULD MAKE PROBABILITY 9/10*8/9 WHICH IS 72/90 AND CLEARLY OVER 1/2. IT SEEMS TO ME THAT THIS IS NOT SUFFICIENT. PERHAPS I AM READING QUESTION WRONG? THANKS.

MC2/10C2 <1/10 --> M(M-1)/10<1/10
M(M-1)<1 this is possible only when M=1 or M=0
So W=9 or 10
p= 9C2/10C2 > 1/2
SUFFCIENT
I AGREE WITH YOU HERE, MAKES PERFECT SENSE.
D

Thanks for pointing out.. thats silly mistake..
I updated the original post.
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05 Sep 2008, 13:43
I get E.

1) As groover pointed out, for W=6

Pr=6/10 * 5/9 = 1/3 < 1/2 thus insuff (it's intuitive that if W6 M4 it's less than 1/2)

2) if M=3

Pr(2M)= 3/10 * 2/9 = 2/30 < 1/10

thus we could have 7 women

Pr(2W) = 7/10 * 6/9 = 7/15 < 1/2

In fact, for M=3, W=7, we have:

Pr(2W)= 7/15
Pr(2M)= 1/15
Pr(1W1M)= 3/10 * 7/9 + 7/10 * 3/9 = 7/15

so E

Where did I go wrong?
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

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05 Sep 2008, 14:05
I get E.

1) As groover pointed out, for W=6

Pr=6/10 * 5/9 = 1/3 < 1/2 thus insuff (it's intuitive that if W6 M4 it's less than 1/2)

2) if M=3

Pr(2M)= 3/10 * 2/9 = 2/30 < 1/10

thus we could have 7 women

Pr(2W) = 7/10 * 6/9 = 7/15 < 1/2

In fact, for M=3, W=7, we have:

Pr(2W)= 7/15
Pr(2M)= 1/15
Pr(1W1M)= 3/10 * 7/9 + 7/10 * 3/9 = 7/15

so E

Where did I go wrong?

you are not wrong buddy!!!

Today is awful day for me!!! busy with work..and meetings.

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Smiling wins more friends than frowning

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Joined: 05 Jul 2008
Posts: 1373

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05 Sep 2008, 14:22
x2suresh wrote:
I get E.

1) As groover pointed out, for W=6

Pr=6/10 * 5/9 = 1/3 < 1/2 thus insuff (it's intuitive that if W6 M4 it's less than 1/2)

2) if M=3

Pr(2M)= 3/10 * 2/9 = 2/30 < 1/10

thus we could have 7 women

Pr(2W) = 7/10 * 6/9 = 7/15 < 1/2

In fact, for M=3, W=7, we have:

Pr(2W)= 7/15
Pr(2M)= 1/15
Pr(1W1M)= 3/10 * 7/9 + 7/10 * 3/9 = 7/15

so E

Where did I go wrong?

you are not wrong buddy!!!

Today is awful day for me!!! busy with work..and meetings.

Arrived at E as well.

Used a slightly diff approach for (2)

The probability that both representatives selected will be men is less than1/10

means

The probability that both representatives selected will NOT be men is greater than or equal to 9/10.

this includes P(WW)+ P(MW) >= 9/10

We cant tell for sure how they will split out. Hence Insufficient.

Any thing wrong with this approach?
Manager
Joined: 09 Jul 2007
Posts: 241

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05 Sep 2008, 14:35
I agree with E.

But, this one is surely a tough question, Option 1 is easily ruled out, but needs a lot of heat and trial for the option 2.

Any easy approach ?
VP
Joined: 05 Jul 2008
Posts: 1373

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05 Sep 2008, 14:47
ssandeepan wrote:
I agree with E.

But, this one is surely a tough question, Option 1 is easily ruled out, but needs a lot of heat and trial for the option 2.

Any easy approach ?

I agree. Did you look at what I did? Is it correct? I arrived at E
Current Student
Joined: 04 Jan 2005
Posts: 282
Location: Milan
Schools: Wharton, LBS, UChicago, Kellogg MMM (Donald Jacobs Scholarship), Stanford, HBS

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05 Sep 2008, 23:07
x2suresh wrote:
you are not wrong buddy!!!

Today is awful day for me!!! busy with work..and meetings.

Eheh, I will tell my nephews that I once corrected x2suresh in math! GMAC should award 20 more points to me for this (my exam is Wednesday)!

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Probability   [#permalink] 05 Sep 2008, 23:07
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If 2 different representatives are to be selected at random

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