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two students are chosen at random

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two students are chosen at random

by arora007 » Fri Aug 13, 2010 3:19 am
If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?
1) There are 50 male students in the class.
2) The probability of selecting one male and one female student is 21/50.

OA is B, but can somebody calculate and show the exact probability
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by kmittal82 » Fri Aug 13, 2010 4:10 am
Clearly (1) is not enough, so lets look at (2)

If you select 2 people, there are only 3 scenarios possible:

1M1F or 2M or 2F

It is already given that the probability for 1M1F = 21/50

Thus, probability for selecting either 2M or 2F = 1 - 21/50 = 29/50

PS: I would've been helpful if the question confirmed that there are only male and female students in the class :)
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by sanju09 » Fri Aug 13, 2010 4:10 am
arora007 wrote:If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?
1) There are 50 male students in the class.
2) The probability of selecting one male and one female student is 21/50.

OA is B, but can somebody calculate and show the exact probability
This question is from papgust's 300 questions!

If the class has m males and f females, then there are (m + f) ^2 ways of selecting two at random with replacement.

Probability that two male students or two female students are selected

= m^2/ (m + f) ^2 + f^2/ (m + f) ^2

= (m^2 + f^2)/ (m + f) ^2.

(1) Only m cannot answer the required probability. Insufficient
(2) This is straight counter probability of the event in stem, which answers the stem as = 1 - 42/50 = 4/25.

Because the probability of selecting one female and one male will be another 21/50. Sufficient

For your satisfaction

Since the probability of selecting one male and one female student is given by m f/ (m + f) ^2, with this known equal to 21/50, we can estimate a ratio for the stem as probability is not more than a proper fraction.

(m^2 + f^2) = (m + f) ^2 - 2 m f = 1 - 2 × 21/50 = 4/25.


[spoiler]B[/spoiler]
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by kmittal82 » Fri Aug 13, 2010 4:20 am
Hi Sanju,

Im a bit confused here, are you treating selection of 1Male and 1Female as a different case to selecting 1Female and 1Male ?
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by sanju09 » Fri Aug 13, 2010 4:21 am
kmittal82 wrote:Hi Sanju,

Im a bit confused here, are you treating selection of 1Male and 1Female as a different case to selecting 1Female and 1Male ?
yes, I am
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by arora007 » Fri Aug 13, 2010 4:30 am
both the answer are right....

kmittal82's is without replacement
sanju09 is with replacement.

the turning point is this step is "sanju09" which was a real trick which i was trying to catch....

"(m^2 + f^2) = (m + f) ^2 - 2 m f = 1 - 2 X 21/50 = 4/25. "

kmittal82: yeah.. it would really have been helpful :) , thats why they say...one should do problems from a good source...lolz..
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by amitakalra88 » Thu Nov 25, 2010 2:14 pm
I can't understand how students can be chosen WITH REPLACEMENT. As in, if the same student is selected twice then the two member committee wont be formed at all, will it?

Also, why are 1M,1F and 1F,1M being treated as separate cases? I cant understand how this is related to replacement. We are taking only combinations here, not permutations, so I don't understand the explanation.

Please help.
Thanks in advance.
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by sanju09 » Thu Dec 02, 2010 2:50 am
amitakalra88 wrote:I can't understand how students can be chosen WITH REPLACEMENT. As in, if the same student is selected twice then the two member committee wont be formed at all, will it?

Also, why are 1M,1F and 1F,1M being treated as separate cases? I cant understand how this is related to replacement. We are taking only combinations here, not permutations, so I don't understand the explanation.

Please help.
Thanks in advance.

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by tomada » Tue Dec 28, 2010 1:39 pm
Specific calculations aside, I guess the point to realize is that the answer can be determined using only Statement (2).
Whether we decide that P(1M + 1F)= 21/50 or 42/50, we know that P(2F or 2M) can be determined.
Time is of the essence.
arora007 wrote:both the answer are right....

kmittal82's is without replacement
sanju09 is with replacement.

the turning point is this step is "sanju09" which was a real trick which i was trying to catch....

"(m^2 + f^2) = (m + f) ^2 - 2 m f = 1 - 2 X 21/50 = 4/25. "

kmittal82: yeah.. it would really have been helpful :) , thats why they say...one should do problems from a good source...lolz..
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by fskilnik@GMATH » Wed Dec 29, 2010 6:35 am
Hi there!

Nice problem... from it I could create a slightly different (pretty malicious) one, below... who would like to give a try? :)

--------------------------------------------------------------------------------------------------------------
From a group of students, two people will be randomly chosen, one after the other, with replacement. What is the probability that two male or two female students will be selected ?

(1) The probability of selecting one male and one female student is 42%
(2) There are twice as many male students as female students
--------------------------------------------------------------------------------------------------------------

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by anshumishra » Wed Dec 29, 2010 7:45 am
fskilnik wrote:Hi there!

Nice problem... from it I could create a slightly different (pretty malicious) one, below... who would like to give a try? :)

--------------------------------------------------------------------------------------------------------------
From a group of students, two people will be randomly chosen, one after the other, with replacement. What is the probability that two male or two female students will be selected ?

(1) The probability of selecting one male and one female student is 42%
(2) There are twice as many male students as female students
--------------------------------------------------------------------------------------------------------------

Regards,
Fabio.

Let me try.

m-> male, f-> female
p(2 male, 2 female) = (m^2+f^2)/(m+f)^2

Statement 1: P(1 m, 1 f) = 42/200 = 21/50 = mf/(m+f)^2
(m^2+f^2)/(m+f)^2 = [(m+f)^2 - 2mf]/(m+f)^2 = 1- 2*21/50 = 4/25 -> Sufficient

Statement 2: m = 2f
=> (m^2+f^2)/(m+f)^2 = 5f^2/9f^2 = 5/9 --> sufficient

Answer D.

[Note : Either I made a mistake or you should change the question so that the probability is same in both cases (to make it more GMATtic]
Last edited by anshumishra on Wed Dec 29, 2010 10:46 am, edited 2 times in total.
Thanks
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by fskilnik@GMATH » Wed Dec 29, 2010 8:55 am
Thanks for your contribution, Anshu!

I said the problem was mallicious, didn´t I?

Rethink sttm (2) in the MOST TRIVIAL scenario...

P.S.: when a certain DS has "D" as its correct answer, each sttm alone does NOT need to give the same answer as the other. I agree that the GREAT MAJORITY of official GMAT DS (D)-answer questions are "coherent" ("harmonious", whatever) but this is NOT always the case and, even if till "today" ALL GMAT DS (D)-answer questions were "coherent", "tomorrow" we could see ACT creating one that is not. Period. (It´s just a matter of reading what (D)-answer means -- and what it does not imply -- many many many times.)
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by anshumishra » Wed Dec 29, 2010 9:43 am
fskilnik wrote:Thanks for your contribution, Anshu!

I said the problem was mallicious, didn´t I?

Rethink sttm (2) in the MOST TRIVIAL scenario...

P.S.: when a certain DS has "D" as its correct answer, each sttm alone does NOT need to give the same answer as the other. I agree that the GREAT MAJORITY of official GMAT DS (D)-answer questions are "coherent" ("harmonious", whatever) but this is NOT always the case and, even if till "today" ALL GMAT DS (D)-answer questions were "coherent", "tomorrow" we could see ACT creating one that is not. Period. (It´s just a matter of reading what (D)-answer means -- and what it does not imply -- many many many times.)
I see, so the "malicious" -> was pointing there. :)
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by fskilnik@GMATH » Wed Dec 29, 2010 10:26 am
anshumishra wrote: I see, so the "malicious" -> was pointing there. :)
As far as the "most trivial scenario" is concerned, I mean it is important to check the case of just 1 female and 2 males, because in this case we would NOT have the possibility of two women selected IF replacement was not allowed.

(I got messed-up while elaborating the question... my intention was an "A" answer, not the "D" that Anshu correctly found.)

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by needthis » Thu Mar 24, 2011 1:07 pm
sanju09 wrote:
arora007 wrote:If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?
1) There are 50 male students in the class.
2) The probability of selecting one male and one female student is 21/50.

OA is B, but can somebody calculate and show the exact probability
This question is from papgust's 300 questions!

If the class has m males and f females, then there are (m + f) ^2 ways of selecting two at random with replacement.

Probability that two male students or two female students are selected

= m^2/ (m + f) ^2 + f^2/ (m + f) ^2

= (m^2 + f^2)/ (m + f) ^2.

(1) Only m cannot answer the required probability. Insufficient
(2) This is straight counter probability of the event in stem, which answers the stem as = 1 - 42/50 = 4/25.

Because the probability of selecting one female and one male will be another 21/50. Sufficient

For your satisfaction

Since the probability of selecting one male and one female student is given by m f/ (m + f) ^2, with this known equal to 21/50, we can estimate a ratio for the stem as probability is not more than a proper fraction.

(m^2 + f^2) = (m + f) ^2 - 2 m f = 1 - 2 × 21/50 = 4/25.


[spoiler]B[/spoiler]
oh dear, probability is my nightmare.
arora007! Can you please enlighten me on why "If the class has m males and f females, then there are (m + f) ^2 ways of selecting two at random with replacement" not (m+f)!/2!(m+f-2)! ways?
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