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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] New post 05 Sep 2013, 08:24
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49% (02:37) correct 51% (01:35) wrong based on 53 sessions
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] New post 05 Sep 2013, 11:18
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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] New post 05 Sep 2013, 21:52
Hi,
Can any one solve this by combination method??

Thanks in Advance,
Rrsnathan
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Re: x, y and z are all unique numbers. If x is chosen randomly [#permalink] New post 05 Sep 2013, 22:16
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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] New post 08 Sep 2013, 02:51
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] New post 08 Sep 2013, 07:23
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] New post 08 Sep 2013, 08:31
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] New post 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 )
Re: x, y and z are all unique numbers. If x is chosen randomly   [#permalink] 09 Sep 2013, 00:39
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