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x, y and z are all unique numbers. If x is chosen randomly [#permalink]
05 Sep 2013, 08:24

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Question Stats:

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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]
08 Sep 2013, 08:31

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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]
09 Sep 2013, 00:39

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]
15 Dec 2014, 00:15

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]
15 Dec 2014, 05:43

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]
16 Dec 2014, 11:51

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]
16 Dec 2014, 21:52

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. _________________

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