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:

x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

05 Sep 2013, 09:24

2

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

54% (02:26) correct
46% (01:46) wrong based on 134 sessions

HideShow timer Statistics

x, y and z are all unique numbers. If x is chosen randomly from the set {7, 8, 9, 10, 11} and y and z are chosen randomly from the set {20, 21, 22, 23}, what is the Probability that x and y are prime and z is not ?

Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

08 Sep 2013, 09:31

1

This post received KUDOS

ankur1901 wrote:

nikhilsehgal wrote:

Probability of X to be Prime is: 2/5 as Only two numbers in the list of {7, 8, 9, 10, 11} i.e 7 and 11.

Probability of Y to be Prime is: 1/4 as Only two numbers in the list of {20, 21, 22, 23} i.e 23

Probability of Z not prime is 3/3 because Not prime numbers in the list of {20, 21, 22, 23} are only three.

Hence X*Y*Z = 2/5 *1/4 *3/3 = 1/10

Hope it helps!!

why the probability of Z is 3/3 and not 3/4. I can choose 3 non prime numbers out of the available 4. Should it not be 3/4?

The question stem states that x, y and z are unique.... _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

09 Sep 2013, 01:39

rrsnathan wrote:

ankur1901 wrote:

nikhilsehgal wrote:

Probability of X to be Prime is: 2/5 as Only two numbers in the list of {7, 8, 9, 10, 11} i.e 7 and 11.

Probability of Y to be Prime is: 1/4 as Only two numbers in the list of {20, 21, 22, 23} i.e 23

Probability of Z not prime is 3/3 because Not prime numbers in the list of {20, 21, 22, 23} are only three.

Hence X*Y*Z = 2/5 *1/4 *3/3 = 1/10

Hope it helps!!

why the probability of Z is 3/3 and not 3/4. I can choose 3 non prime numbers out of the available 4. Should it not be 3/4?

Hi,

Y has been picked from the set before Z. So there is only 3 numbers in the list. hence the probablity is Z =3/3

Regards, Rrsnathan.

Just one quick question...how do we know that y is picked up before z...the question says that y & z are picked up randomly...i assumed that they are picked up simultaneously....if we dont know the order, the combination will be (1C1 * 3C1)/4C2 ; Irrespective we get E as the answer (2/5 * 1/4 =1/10 )

x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

15 Dec 2014, 01:15

1

This post received KUDOS

Expert's post

bagdbmba wrote:

x, y and z are all unique numbers. If x is chosen randomly from the set {7, 8, 9, 10, 11} and y and z are chosen randomly from the set {20, 21, 22, 23}, what is the Probability that x and y are prime and z is not ?

A. 1/5 B. 3/20 C. 13/20 D. 3/10 E. 1/10

x from {7, 8, 9, 10, 11} - there are 2 primes here 7 and 11. Probability that x is prime is 2/5.

y and z from {20, 21, 22, 23} - there is only one prime number here 23. Probability that y is prime is 1/4. Now you are left with 3 numbers since y and z must be different. You can select z in any way out of those 3 numbers since all are non prime. So probability that z is not prime is 1.

Probability that x and y are prime but z is not = (2/5)*(1/4)*1 = 2/20 = 1/10

Answer (E)

Note that out of y and z, it doesn't matter what you pick first. If you pick z first, the probability that it is not prime is (3/4). Now you are left with 3 numbers one of which is prime. The probability that y is prime is (1/3). So probability that y is prime and z is not = (3/4)*(1/3) = (1/4) - same as before _________________

x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

15 Dec 2014, 06:43

VeritasPrepKarishma wrote:

bagdbmba wrote:

x, y and z are all unique numbers. If x is chosen randomly from the set {7, 8, 9, 10, 11} and y and z are chosen randomly from the set {20, 21, 22, 23}, what is the Probability that x and y are prime and z is not ?

A. 1/5 B. 3/20 C. 13/20 D. 3/10 E. 1/10

x from {7, 8, 9, 10, 11} - there are 2 primes here 7 and 11. Probability that x is prime is 2/5.

y and z from {20, 21, 22, 23} - there is only one prime number here 23. Probability that y is prime is 1/4. Now you are left with 3 numbers since y and z must be different. You can select z in any way out of those 3 numbers since all are non prime. So probability that z is not prime is 1.

Probability that x and y are prime but z is not = (2/5)*(1/4)*1 = 2/20 = 1/10

Answer (E)

Note that out of y and z, it doesn't matter what you pick first. If you pick z first, the probability that it is not prime is (3/4). Now you are left with 3 numbers one of which is prime. The probability that y is prime is (1/3). So probability that y is prime and z is not = (3/4)*(1/3) = (1/4) - same as before

Nice! Excellent explanation on picking z vs. y first. Thank you.

Can you please expound on when/whether we have to subtract the probability of the opposite condition? For example, translating the math to a condition: Picking x out of the first set so it is prime, and picking y and z out of the second set such that y is prime, and z is not. Do we not have to subtract the possibility that z is prime, or is that implied given that our condition is X is prime AND y is prime AND z is not prime, and given that z cannot be prime if y is prime? Would we have to subtract the possibility if two prime numbers existed in the second set?

Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

16 Dec 2014, 12:51

VeritasPrepKarishma wrote:

bagdbmba wrote:

x, y and z are all unique numbers. If x is chosen randomly from the set {7, 8, 9, 10, 11} and y and z are chosen randomly from the set {20, 21, 22, 23}, what is the Probability that x and y are prime and z is not ?

A. 1/5 B. 3/20 C. 13/20 D. 3/10 E. 1/10

x from {7, 8, 9, 10, 11} - there are 2 primes here 7 and 11. Probability that x is prime is 2/5.

y and z from {20, 21, 22, 23} - there is only one prime number here 23. Probability that y is prime is 1/4. Now you are left with 3 numbers since y and z must be different. You can select z in any way out of those 3 numbers since all are non prime. So probability that z is not prime is 1.

Probability that x and y are prime but z is not = (2/5)*(1/4)*1 = 2/20 = 1/10

Answer (E)

Note that out of y and z, it doesn't matter what you pick first. If you pick z first, the probability that it is not prime is (3/4). Now you are left with 3 numbers one of which is prime. The probability that y is prime is (1/3). So probability that y is prime and z is not = (3/4)*(1/3) = (1/4) - same as before

Hi Karishma , I have a doubt here. Even though picking x first or picking y first gives us equal probability i.e 1/4 resulting in probability of 1/10. Ideally shouldn't we consider the both cases i.e.picking x first or picking y first while calculating the final probability? In that case final probability should be 1/10+1/10 = 1/5 Please correct me if I am wrong.

Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

16 Dec 2014, 22:52

Expert's post

solitaryreaper wrote:

VeritasPrepKarishma wrote:

bagdbmba wrote:

x, y and z are all unique numbers. If x is chosen randomly from the set {7, 8, 9, 10, 11} and y and z are chosen randomly from the set {20, 21, 22, 23}, what is the Probability that x and y are prime and z is not ?

A. 1/5 B. 3/20 C. 13/20 D. 3/10 E. 1/10

x from {7, 8, 9, 10, 11} - there are 2 primes here 7 and 11. Probability that x is prime is 2/5.

y and z from {20, 21, 22, 23} - there is only one prime number here 23. Probability that y is prime is 1/4. Now you are left with 3 numbers since y and z must be different. You can select z in any way out of those 3 numbers since all are non prime. So probability that z is not prime is 1.

Probability that x and y are prime but z is not = (2/5)*(1/4)*1 = 2/20 = 1/10

Answer (E)

Note that out of y and z, it doesn't matter what you pick first. If you pick z first, the probability that it is not prime is (3/4). Now you are left with 3 numbers one of which is prime. The probability that y is prime is (1/3). So probability that y is prime and z is not = (3/4)*(1/3) = (1/4) - same as before

Hi Karishma , I have a doubt here. Even though picking x first or picking y first gives us equal probability i.e 1/4 resulting in probability of 1/10. Ideally shouldn't we consider the both cases i.e.picking x first or picking y first while calculating the final probability? In that case final probability should be 1/10+1/10 = 1/5 Please correct me if I am wrong.

It doesn't matter in what way you pick y and z, if the end result is the same, it needs to be considered a single case. _________________

Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

Show Tags

19 Feb 2016, 09:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...