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# If 2 different representatives are to be selected at random

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Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 243

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15 Oct 2009, 16:01

we need to find if P(W)=p is > 1/2 or not

Prob of both being women = 1 - {p(of two men) + p(of 1 M , 1W)}

p = 1 - (x+y)

S1 ) Sufficient - from this we can tell if p is > 1/2 or not
because assuming 6 w 4m, 7w 3m, etc.,
we can calculate x and y and tell is p>1/2 or not

S2 ) Insufficient - coz we are given multiple values for x, but no way to know y
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Senior Manager
Joined: 31 Aug 2009
Posts: 369
Location: Sydney, Australia

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15 Oct 2009, 16:30
(1) More than 1/2 of the 10 employees are women.
Assuming 9 out of 10 are women. Probability = 9/10 * 8/9 = 72/90 This is > 1/2
Assuming 7 out of 10 are women. Probability = 7/10 * 6/9 = 42/90 This is < ½
Insufficient.

(2) The probability that both representatives selected will be men is less than 1/10?
P(both women) = 1- [p(both men) + p(1 man, 1 woman)]
This does not tell us anything about the number of women.

1 and 2 Together. Still insuff as the number of women is still variable.
ANS = E
Math Expert
Joined: 02 Sep 2009
Posts: 55271

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15 Oct 2009, 17:33
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?

What is the probability of choosing 2 women out of 10 people w/10*(w-1)/9 and this should be >1/2. So we have w/10*(w-1)/9>1/2 --> w(w-1)>45 this is true only when w>7. (w # of women <=10)

So basically question asks is w>7

(1) w>5 not sufficient
(2) (10-w)/10*(10-w-1)/9<1/10 --> (10-w)(9-w)<9 --> w>6 not sufficient

(1)+(2) w>5, w>6 not sufficient

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Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 243

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15 Oct 2009, 19:21
Thanks Yangsta and Buniel

u guys opened my eyes thx
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Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip
Manager
Joined: 19 Nov 2007
Posts: 171

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07 Nov 2009, 06:01
According Stem 1

Women>5
The probability of two women being selected could be 7*6/90 = 0.46 or 8*7/90 =0.62 etc…; Conflicting results So A and D are out

According to the stem 2
If x is the number of men, then the probability of two men being selected is x(x-1)/90 and x(x-1)/90 <1/10 =>x =2 or 3 (x being a integer)
So the number of women could be 7 or 8

If there are 7 women then probability of two women being selected is 7*6/90=0.46 or 8*7/90=0.62 probability of two women so D is out.

If we combine the two stems it is equivalent to the second stem. So choice D is out

Manager
Joined: 17 Aug 2009
Posts: 167

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09 Dec 2009, 09:08
Statement 1 = Insufficient

If Women are 7,

7/10 * 6/9 = 42/90 which Is less than ½

If women are 8,

8/10 * 7/9 which is more than half

Two results. Therefore Insufficient

Statement 2------------Insufficient

Since it is mentioned that the probability that both will be men is less than 1/10

Men can be 3 or 2, since

3/10 * 2/9 and 2/10* 1/9 are both less than 1/10

Which leaves us with women being 7 or 8 (in which case it could be greater than ½ or less than ½)

Taking both together,

Leaves us with the same solution

Therefore E
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Re: If 2 different representatives are to be selected at random  [#permalink]

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26 Aug 2017, 22:52
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Re: If 2 different representatives are to be selected at random   [#permalink] 26 Aug 2017, 22:52

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