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

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

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05 Sep 2013, 09:24
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54% (02:26) correct 46% (01:46) wrong based on 134 sessions

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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 ?

A. 1/5
B. 3/20
C. 13/20
D. 3/10
E. 1/10
[Reveal] Spoiler: OA

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

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05 Sep 2013, 12:18
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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!!
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Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

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05 Sep 2013, 22:52
Hi,
Can any one solve this by combination method??

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

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05 Sep 2013, 23:16
1
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Total number of outcomes = (number of ways x can be chosen) * (Number of ways y and z could be chosen)= 5c1 *(4C1*3C1)

Number of ways x is chosen and it is prime = 2C1
Number of ways y is chosen and it is prime = 1C1
Number of ways z is chosen and it is not-prime = 3C1

P = 2C1*1C1*3C1/ 5C1*(4C1*3C1) = 1/10

Hope that helps.

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

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08 Sep 2013, 03:51
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?
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Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

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08 Sep 2013, 08:23
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.
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Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

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08 Sep 2013, 09:31
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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....
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Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink]

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

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14 Dec 2014, 11:53
Anyone able to explain this one further?
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x, y and z are all unique numbers. If x is chosen randomly [#permalink]

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15 Dec 2014, 01:15
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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

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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 14 Nov 2014 Posts: 108 Location: United States GMAT 1: 740 Q49 V41 GPA: 3.34 WE: General Management (Aerospace and Defense) Followers: 1 Kudos [?]: 28 [0], given: 0 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? Thanks for the help. Manager Joined: 23 Sep 2013 Posts: 104 Concentration: Strategy, Marketing WE: Engineering (Computer Software) Followers: 1 Kudos [?]: 6 [0], given: 64 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6480 Location: Pune, India Followers: 1760 Kudos [?]: 10498 [0], given: 206 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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Re: x, y and z are all unique numbers. If x is chosen randomly   [#permalink] 19 Feb 2016, 09:06
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