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!
two students are chosen at random
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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
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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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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yes, I amkmittal82 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 ?
The mind is everything. What you think you become. -Lord Buddha
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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..
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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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.
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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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.
over to arora007...
The mind is everything. What you think you become. -Lord Buddha
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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.
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..
I'm really old, but I'll never be too old to become more educated.
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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.
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.
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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
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
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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 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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I see, so the "malicious" -> was pointing there.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.)
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
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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.anshumishra wrote: I see, so the "malicious" -> was pointing there.
(I got messed-up while elaborating the question... my intention was an "A" answer, not the "D" that Anshu correctly found.)
Regards,
Fabio.
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oh dear, probability is my nightmare.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]
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?