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# If 2 different representatives are to be selected at random

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If 2 different representatives are to be selected at random [#permalink]  22 Aug 2006, 23:09
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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 5 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10 .
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[#permalink]  22 Aug 2006, 23:20
E

Let w = women
m = men

St1: least prob is 6/10 * 5/9 = 1/3 and max prob is 1.: INSUFF

St2: m/10 * (m-1)/9 < 1/10
i.e m * (m-1) < 9
i.e m are either 0,1,2 or 3.
So women are greater than 6.
least prob is 7/10 * 6/9 = 1/15 max prob is 1: INSUFF

Together:
We get women are greater than 6. Same as statement 2.
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[#permalink]  22 Aug 2006, 23:25
yup as usual your right on the money
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[#permalink]  23 Aug 2006, 08:11
ps_dahiya wrote:
E

St1: least prob is 6/10 * 5/9 = 1/3 and max prob is 1.: INSUFF

in this particular step, I always get confused.
I dont know what rule to apply to get probability of selecting 2 women.
Choice 1:
Like you have done probablility = 6/10 * 5/9
Choice 2: (what I did when trying to solve this problem on my own!)
Least Probability = 1/6*1/5
(out of 6 women, probability of selecting the first is 1/6. now that one is selected, the probability of selecting the next is 1 out of the 5 remaining).

what i would like to know is why you have taken choice 1 and not choice 2. is there a particular formula / method that indicates when to use what pls ?
any help on this would be highly appreciated.
thanks.
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Prashrash.

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[#permalink]  23 Aug 2006, 14:08
prashrash wrote:
ps_dahiya wrote:
E

St1: least prob is 6/10 * 5/9 = 1/3 and max prob is 1.: INSUFF

in this particular step, I always get confused.
I dont know what rule to apply to get probability of selecting 2 women.
Choice 1:
Like you have done probablility = 6/10 * 5/9
Choice 2: (what I did when trying to solve this problem on my own!)
Least Probability = 1/6*1/5
(out of 6 women, probability of selecting the first is 1/6. now that one is selected, the probability of selecting the next is 1 out of the 5 remaining).

what i would like to know is why you have taken choice 1 and not choice 2. is there a particular formula / method that indicates when to use what pls ?
any help on this would be highly appreciated.
thanks.

Choice 2 is wrong.

I simply ask: What is the prob of selecting a women from a group of 6 women and 4 men. It will be 6/10.
Now we have selected a women there are 5 women remaining and total of 9 people remaining. Prob of selecting a women from a group of 5 women and 4 men. It is 5/9
So prob that first one is women and second one is also women = 6/10 * 5/9

Hope this helps.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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[#permalink]  24 Aug 2006, 05:32
prashrash wrote:
Choice 2: (what I did when trying to solve this problem on my own!)
Least Probability = 1/6*1/5
(out of 6 women, probability of selecting the first is 1/6. now that one is selected, the probability of selecting the next is 1 out of the 5 remaining).

Choice 2 come sinto play when you want to select a particular woman from that group.... Lest say you had to choose Jenny and Katie both... then you can used Choice 2...
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[#permalink]  24 Aug 2006, 08:13
Thanks guys. Appreciate the explanations. Yes, they have helped me to a great extent.
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Cheers!
Prashrash.

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[#permalink]  25 Aug 2006, 00:02
ps_dahiya wrote:
sumitsarkar82 wrote:
prashrash wrote:
Choice 2: (what I did when trying to solve this problem on my own!)
Least Probability = 1/6*1/5
(out of 6 women, probability of selecting the first is 1/6. now that one is selected, the probability of selecting the next is 1 out of the 5 remaining).

Choice 2 come sinto play when you want to select a particular woman from that group.... Lest say you had to choose Jenny and Katie both... then you can use Choice 2...

That is so correct.

Learnt it from you sir
[#permalink] 25 Aug 2006, 00:02
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# If 2 different representatives are to be selected at random

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