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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 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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New post 21 Jun 2008, 04:50
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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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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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.
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New post 20 Mar 2011, 09:16
Onell 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.


Statement 1 only tells us about number of male students but we do not knpw number of females and hence cant calculate probabilities - Insufficient

Statement 2 tells us the probability of selecting one male and one female student and hence probability of either both male or both female is = 1-P(one male and one female). Hence, Sufficient.

Answer B
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New post 24 Jul 2014, 09:50
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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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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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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New post 25 Jul 2014, 08:50
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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New post 25 Jul 2014, 10:14
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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New post 25 Jul 2014, 21:11
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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New post 26 Jul 2014, 07:25
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]

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New post 26 Jul 2014, 08:44
Thanks a lot for explanation
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Re: If two students are chosen at random with replacement from a [#permalink]

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New post 14 Apr 2016, 23:38
Bunuel wrote:
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.


I chose C as the answer. How do we know that the ratio of F/Class is not 42/100 or 126/150? The final answer chants as a result. Unless we know the number of students in the class is 50, we don't know where we might be going wrong.

Can someone please correct me if I'm wrong?
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New post 15 Apr 2016, 00:30
Avinashs87 wrote:
Bunuel wrote:
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.


I chose C as the answer. How do we know that the ratio of F/Class is not 42/100 or 126/150? The final answer chants as a result. Unless we know the number of students in the class is 50, we don't know where we might be going wrong.

Can someone please correct me if I'm wrong?


The probability of an outcome is the ratio of favorable outcomes to the total number of outcomes. It does not matter whether the ratio itself is reduced to its lowest term or not. So, for example, the probability of 1/3, 2/6, 3/9, ... are the same probabilities. We need the probability of blue events, which is 29/50. You can write it as 58/100 or 290/500, it won't change the numerical value of the ratio.
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Re: If two students are chosen at random with replacement from a [#permalink]

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New post 15 Apr 2016, 01:43
The probability of an outcome is the ratio of favorable outcomes to the total number of outcomes. It does not matter whether the ratio itself is reduced to its lowest term or not. So, for example, the probability of 1/3, 2/6, 3/9, ... are the same probabilities. We need the probability of blue events, which is 29/50. You can write it as 58/100 or 290/500, it won't change the numerical value of the ratio.[/quote]

Understand...but let me ask the question in a different way. How do we know that 21/50 is not a ratio in its reduced form?
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New post 15 Apr 2016, 01:49
Avinashs87 wrote:
The probability of an outcome is the ratio of favorable outcomes to the total number of outcomes. It does not matter whether the ratio itself is reduced to its lowest term or not. So, for example, the probability of 1/3, 2/6, 3/9, ... are the same probabilities. We need the probability of blue events, which is 29/50. You can write it as 58/100 or 290/500, it won't change the numerical value of the ratio.


Understand...but let me ask the question in a different way. How do we know that 21/50 is not a ratio in its reduced form?[/quote]

How does this matter? The probability is a number, which we are given to be 21/50. It does not matter how it's written.
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If two students are chosen at random with replacement from a [#permalink]

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New post 15 Apr 2016, 01:56
Avinashs87 wrote:
The probability of an outcome is the ratio of favorable outcomes to the total number of outcomes. It does not matter whether the ratio itself is reduced to its lowest term or not. So, for example, the probability of 1/3, 2/6, 3/9, ... are the same probabilities. We need the probability of blue events, which is 29/50. You can write it as 58/100 or 290/500, it won't change the numerical value of the ratio.

Understand...but let me ask the question in a different way. How do we know that 21/50 is not a ratio in its reduced form?


Hi,

21/50 is a ratio in its reduced form and cannot be divided further. 21 and 50 are co primes.

A Probability of 21/50 is same as the probability of 42/100. The question is concerned only on the probability aspect and not on the number of students.

The probability that 1 event occurs out of a total of 2 events is same as 2 events occurring out of a total of 4 events.
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