Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If two students are chosen at random with replacement from a [#permalink]

Show Tags

21 Jun 2008, 03:02

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

51% (01:53) correct
49% (01:12) wrong based on 166 sessions

HideShow timer Statistics

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.

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

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.

If two students are chosen at random with replacement from a [#permalink]

Show Tags

25 Jul 2014, 03:23

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

25 Jul 2014, 08:03

1

This post received KUDOS

Expert's post

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

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

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

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

25 Jul 2014, 10:14

Expert's post

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. _________________

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

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

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

26 Jul 2014, 07:25

Expert's post

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) _________________

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

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.

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

15 Apr 2016, 00:30

Expert's post

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. _________________

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

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?

Re: If two students are chosen at random with replacement from a [#permalink]

Show Tags

15 Apr 2016, 01:49

Expert's post

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. _________________

If two students are chosen at random with replacement from a [#permalink]

Show Tags

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.

gmatclubot

If two students are chosen at random with replacement from a
[#permalink]
15 Apr 2016, 01:56

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...