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If two students are chosen at random with replacement from a

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If two students are chosen at random with replacement from a [#permalink] New post 21 Jun 2008, 02:02
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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.
[Reveal] Spoiler: OA
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Re: If two students are chosen at random with replacement from a [#permalink] New post 21 Jun 2008, 03:50
Quote:
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.


I’ve got B for this one. Here is how:

Statement 1: Clearly insuff. There may be 100 female as well as 0, which will produce very different answers.

Statement 2. Let n be the number of students in the class, m be the number of male students. Then the number of females is (n-m). So, this statement gives us
p = 2*p(one male) * p(one female) = 2*m/n * (n-m)/n = 2*(nm – m^2)/n^2. And this equals 21/50.

Here the most interesting part begins. At first glance, we have one eq. with two variables. And Statement 1 gives us the value for one of the variables. So, at this stage, it might be tempting to select C. But this would be a mistake.

Why? Now look again at what exactly is asked: p(two males or two females) = p(two males) + p(two females) = m^2/n^2 + (n-m)^2/n^2 = (n^2 + 2m^2 – 2nm)/n^2 = (n^2 – 2(nm-m^2))/n^2 = 1-2*(nm-m^2)/n^2. We need not to now what m and n are, we need to now a relation between them. And Statement 2 gives us exactly the value of the required expression 2(nm-m^2)/n^2 = 21/50.

So, 2) was sufficient.

****
Edited later:
I’ve thought of alternative approach to this problem – an easy one.
Statement 2:
1 = p(two males) + p(two females) + p(one male and one female). Statement 2 gives us p(one male and one female).
Thus, we know p(two males) + p(two females). => 2 is suff.
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Re: If two students are chosen at random with replacement from a [#permalink] New post 22 Jun 2008, 21:36
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It is a direct B:

Using P(A')=1-P(A)

St1: Insuffi
St2:We directly have the ans as: 1-21/50 = 29/35
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Re: If two students are chosen at random with replacement from a [#permalink] New post 24 Jul 2014, 08:50
Cartic,

Can you please explain how it is "P(A')=1-P(A)"? Got the correct ans though but unable to understand the approach. Thanks in advance
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If two students are chosen at random with replacement from a [#permalink] New post 25 Jul 2014, 02:23
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sri30kanth 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.

Cartic,

Can you please explain how it is "P(A')=1-P(A)"? Got the correct ans though but unable to understand the approach. Thanks in advance


There are only 4 cases when you are selecting 2 people:
1. (male, male);
2. (female, female);
3. (male, female);
4. (female, male).

The sum of the probabilities of those 4 cases is 1. We need the sum of the probabilities of the first 2 cases: P(1 + 2). The second statement says that the sum of the probabilities of the last 2 cases is 21/50: P(3 + 4) = 21/50. therefore P(1 + 2) = 1 - P(3 + 4) = 1 - 21/50 = 29/50.

Generally, P(x) = 1 - P(not x).

Hope it's clear.
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Re: If two students are chosen at random with replacement from a [#permalink] New post 25 Jul 2014, 07:03
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lexis wrote:
If two students are chosen at randomwith replacementfrom 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.


A Detailed Explanation

Question : probability that two male students or two female students are selected?

Statement 1)There are 50 male students in the class.
Male = 50 But No. of Female Students is unknown therefore Probability can't be calculated
INSUFFICIENT

Statement 2) The probability of selecting one male and one female student is 21/50
i.e. (No. of Males/Male+female) x (No. of Female/Male+female)x2! = 21/50

BUT since we are calculating probability with replacement therefore

Probability of a male selected = 1-probability of a female selected

therefore, If probability of one male selected = x
then probability of one female selected = 1-x

therefore, x(1-x)*2 = 21/50
i.e. x(1-x) = 21/100
i.e. x = 3/10 or x = 7/10

i.e. out of 10 students,
Case 1: male = 3 and female = 7
probability of both male = (3/10)x(3/10) = 9/100
probability of both Female = (7/10)x(7/10) = 49/100
i.e. Probability of Both male or Both Female = (9/100)+(49/100) = 58/100 = 29/50


Case 2: male = 7 and female = 3
probability of both Female = (3/10)x(3/10) = 9/100
probability of both Male = (7/10)x(7/10) = 49/100
i.e. Probability of Both male or Both Female = (9/100)+(49/100) = 58/100 = 29/50

Either way we get unique answer therefore, SUFFICIENT

Answer: Option
[Reveal] Spoiler:
B

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Re: If two students are chosen at random with replacement from a [#permalink] New post 25 Jul 2014, 07:50
Thanks Cartic for explanation. But shouldn't we multiply the probability by 2? That is 1 -(2*21/50) because we have to subtract probability of both one male and one female and one female and one male. Please clarify. Thanks again
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Re: If two students are chosen at random with replacement from a [#permalink] New post 25 Jul 2014, 09:14
sri30kanth wrote:
Thanks Cartic for explanation. But shouldn't we multiply the probability by 2? That is 1 -(2*21/50) because we have to subtract probability of both one male and one female and one female and one male. Please clarify. Thanks again


The probability of (one male and one female) and (one female and one male) have already been taken care of in 21/50 therefore it doesn't need to be subtracted.

Perhaps it will be clearer if you could check the detailed explanation given above.
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Re: If two students are chosen at random with replacement from a [#permalink] New post 25 Jul 2014, 20:11
Thank you GMAT insight for the insight to the problem :) But can u throw some light on why we are multiplying with 2! (No. of Males/Male+female) x (No. of Female/Male+female)x2! ? Thanks again
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Re: If two students are chosen at random with replacement from a [#permalink] New post 26 Jul 2014, 06:25
sri30kanth wrote:
Thank you GMAT insight for the insight to the problem :) But can u throw some light on why we are multiplying with 2! (No. of Males/Male+female) x (No. of Female/Male+female)x2! ? Thanks again


Hi Srikanth,

Probability = Favorable Outcomes / Total Outcomes

Fact 1) We can calculate Favorable Outcomes and Total Outcomes both Using Combination (Selection Technique nCr) or Both Using Permutation (Arrangement Technique Using nPr or nCr x r! or r x (r-1) x (r-2) x ...3x2x1)

e.g. Probability of 2 cards picked from a pack of 52 cards being "Kings" is

Either 4C2 / 52C2 [Using Combination] OR 4C2 x 2! / 52C2x2! OR 4x3 / 52x51

[I hope you can see that nCr x r! includes arrangements of 2 picked card i.e. which king is picked first from the pack of card and which one is second]


Regarding your Question:

I have calculated the probability by using the arrangement
i.e. Probability of taking one boy (3/10) and then probability of taking one girl (7/10) but since first selection could be a girl followed by second selection of a boy therefore we are taking arrangement into account therefore Multiply by 2! (because arrangement of 2 selected persons can be done in 2! ways)
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Re: If two students are chosen at random with replacement from a [#permalink] New post 26 Jul 2014, 07:44
Thanks a lot for explanation
Re: If two students are chosen at random with replacement from a   [#permalink] 26 Jul 2014, 07:44
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