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Director  Joined: 03 Sep 2006
Posts: 767
Re: gmatprep combs and probs ds  [#permalink]

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Case ( i )

Women can be = 6, 7, 8, 9, 10 and Men respectively will be 4,3,2,1,0.

Say, W = 6 and M = 4

Probability that both representatives selected will be
women, is p

p = 6C2/10C2 = 15/45 = 1/3

1/3 < 1/2

Say, W =10, M =0

p =1 > 1/2

So we can't answer.

Case ( ii )

Probability that both representatives selected will be
men, is P < 1/10

Probability that None are men = 1-P = Probability that both are women

p<1/10

1-p = 1-1/10 = 9/10 > 1/2

Thus ( ii ) answers.

Answer should be "B"
VP  Joined: 22 Nov 2007
Posts: 1026
Re: gmatprep combs and probs ds  [#permalink]

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LM wrote:
Case ( i )

Women can be = 6, 7, 8, 9, 10 and Men respectively will be 4,3,2,1,0.

Say, W = 6 and M = 4

Probability that both representatives selected will be
women, is p

p = 6C2/10C2 = 15/45 = 1/3

1/3 < 1/2

Say, W =10, M =0

p =1 > 1/2

So we can't answer.

Case ( ii )

Probability that both representatives selected will be
men, is P < 1/10

Probability that None are men = 1-P = Probability that both are women

p<1/10

1-p = 1-1/10 = 9/10 > 1/2

Thus ( ii ) answers.

Answer should be "B"

you miss the case in which one is man and the other is woman.

this is the fastest way I have found to solve this one:

we must show whether wc2/10c2 >1/2

1. w>5
let's try with 6 or 8. with 6 we have 1/3, less than 1/2. with 8 we have 28/45, >1/2. not suff
2.mc2/10c2<1/10, mc2<4.5, m can be 3,2,1 or 0. if m=3, w=7, 7c2/10c2<1/2 ok. if m=2 and w=8, always 28/45>1/2 not suff

combining doesn't say much further. OA is E
Manager  Joined: 04 Jan 2008
Posts: 108
Zumit DS 009  [#permalink]

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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 than 1/10

Originally posted by dancinggeometry on 06 Sep 2008, 00:21.
Last edited by dancinggeometry on 06 Sep 2008, 00:50, edited 1 time in total.
Director  Joined: 23 Sep 2007
Posts: 725
Re: Zumit DS 009  [#permalink]

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dancinggeometry 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 21 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10

(1) More than 21 of the 10 employees are women.???
Senior Manager  Joined: 10 Mar 2008
Posts: 301
PS: Probability  [#permalink]

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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 than 1/10

Originally posted by vksunder on 16 Sep 2008, 14:26.
Last edited by vksunder on 17 Sep 2008, 12:13, edited 1 time in total.
Manager  Joined: 31 Jul 2008
Posts: 242
Re: PS: Probability  [#permalink]

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something wrong with the question (p can never be greater than 1)
VP  Joined: 21 Jul 2006
Posts: 1361
Re: PS: Probability  [#permalink]

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vksunder 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 than 1/10

w= # of women
m= # of men

Probability that both selected are women:

$$(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2$$, so the question is asking us whether $$w^2-w>45$$--->because $$(1/2)*90 = 45$$

(1) So:
if $$w=6$$, then $$6^2-6 < 45$$ False
if $$w=9$$, then $$9^2-9 > 45$$ True --->NOT SUFF.

(2) $$(m/10)*((m-1)/9) = ((m^2-m)/90) < 10%*90$$ -----> $$(m^2-m/90) < (9/90)$$, so $$m^2-m < 9$$

pick numbers: if m=2 then w=8, and if m=3 then w=7.

so when $$w=8$$---> $$8^2- 8 >45$$ true
so when $$w=7$$---> $$7^2 - 7 <45$$ False ----> NOT Suff.

(Together)

$$W=8$$, then $$8^2-8 >45$$, and when $$w=7$$, then $$7^2-7 < 45$$

Answer E.

I just edited my calculations once you've corrected your typo.

Originally posted by tarek99 on 16 Sep 2008, 15:47.
Last edited by tarek99 on 17 Sep 2008, 14:46, edited 4 times in total.
Senior Manager  Joined: 10 Mar 2008
Posts: 301
Re: PS: Probability  [#permalink]

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Sorry about the typo. I've replaced 21 with the correct value. That is, 1/2
VP  Joined: 21 Jul 2006
Posts: 1361
Re: PS: Probability  [#permalink]

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I just corrected my calculations after you've corrected your typo. Answer should be E. What's the OA?
Director  Joined: 30 Jun 2008
Posts: 968
Re: PS: Probability  [#permalink]

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tarek99 wrote:

$$(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2$$, so the question is asking us whether $$w^2-w>45$$--->because $$(1/2)*90 = 45$$

tarek .. I do not get this step(which happens to be the basis for this problem)

Please elaborate ..
_________________
"You have to find it. No one else can find it for you." - Bjorn Borg
Manager  Joined: 28 Aug 2008
Posts: 91
Re: PS: Probability  [#permalink]

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Quote:
(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2, so the question is asking us whether w^2-w>45--->because (1/2)*90 = 45
Quote:

Amitdgr... let me attempt to elaborate here:

Tarek's formula is just the words from the stem as an equation.. for example

From 1, say 6/10 are female... this means that when the first person is selceted the prob. is 6/10 (thus x/10) , when the next person is selceted, the prob. of Female becomes 5/9 (hence x-1/9), the total probability is 6/10 * 5/9 = 1/3 where

(w/10)*((w-1)/9) = (w^2-w)/90

is equal to (6/10)*((6-1)/9) = 30/90
Manager  Joined: 28 Apr 2008
Posts: 108
Re: Zumit DS 009  [#permalink]

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My answer is E

lets say there are 3 men and 7 woman-

the chance for 2 men is 3/10*2/9=6/90 <1/10

the chance for 2 woman is 7/10*6/9 = 42/90=p.....p < 1/2

lets say thre are 2 men and 8 woman-

the chance for 2 men is 2/10*1/9=2/90= <1/10

the chance for 2 woman is 8/10*7/9 = 56/90=p ..... p > 1/2

so must be E
VP  Joined: 17 Jun 2008
Posts: 1162
Re: Zumit DS 009  [#permalink]

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dancinggeometry 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 than 1/10

IMO D
1)INSUFFI ,say women=5 then 5*4/10*9 <1/2

when women=8 then p>1/2

2)INSUFFI thouh im not able to put down he calculation

IMO E
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Manager  Joined: 06 Jul 2007
Posts: 207
Re: still prob  [#permalink]

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marcodonzelli 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 than 1/10.

Fact 1: p(w) > 1/2

Let there are n women out of 10 employees. so n > 5.

number of ways of selecting 2 women out of n = nC2 = n*(n-1)/2.

total number of possible ways of selecting 2 employees out of 10 = 10C2 = 45.

probability that two selected employees are women = n(n-1)/90

n = 6, p=1/3
n = 7, p=42/90
n = 8, p=56/90.

1st is not sufficient.

Fact 2: let number of men in the sample is m. number of ways of selecting 2 men out of m = mC2 = m*(m-1)/2.

total number of ways = 45.

p(both men) = m*(m-1)/90

m*(m-1)/90 < 1/10

m*(m-1)<9
m = 3, 2, 1

second not sufficient.

combining both, we have two solutions 8 women and 2 men OR 7 women or 3 men.

Hence E.
Intern  Joined: 21 Jul 2006
Posts: 17
Re: still prob  [#permalink]

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Is this question correct?
p is the probability?
How can "p>21"? Should not p be between 0 and 1?
Manager  Joined: 06 Jul 2007
Posts: 207
Re: still prob  [#permalink]

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matrix777 wrote:
Is this question correct?
p is the probability?
How can "p>21"? Should not p be between 0 and 1?

I'm not sure what question you are looking at. this question asks "is p>1/2"?
Intern  Joined: 18 Jul 2009
Posts: 37
Probability problems  [#permalink]

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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 than 1/10

Thanks in advance
Intern  Joined: 24 Jun 2009
Posts: 34
Re: Probability problems  [#permalink]

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E

A is insuff
B is insuff as it provides info for not two men i.e. it could either be two men or one man and one woman
together the prob of 2 women is between 1/3 and 9/10. therefore insuff.

What is the OA?
Manager  Joined: 14 Aug 2009
Posts: 115
Re: Probability problems  [#permalink]

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hrish88 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 than 1/10

Thanks in advance

for (1), suppose W=6, then both representatives are women would be 6C2, and total select ways are 10C2
p=6C2/10C2=(6*5)/(10*9)=1/3<1/2
therefore (1) is nsf.

for (2) suppose there are M mem,
then Pman=MC2/10C2<1/10, so, M(M-1)<9, M=0, 1, 2, 3
for M=3, then W=7,
p=7C2/10C2=(7*6)/(10*9)=42/90<45/90=1/2

therefore answer is E
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Kudos me if my reply helps!
Intern  Joined: 24 Jun 2009
Posts: 34
Re: Probability problems  [#permalink]

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The condition II states that the prob of selecting two men is less than 1/10. This provides information for "prob of not selecting two men" and this means that either both reps are women OR one rep is a man and one rep is a woman. This is insuff coz it does not help calculate the prob of two women reps >1/2. hope this helps! Re: Probability problems   [#permalink] 19 Aug 2009, 05:50

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