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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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49% (01:33) wrong based on 51 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 ?

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

gmatclubot

Re: x, y and z are all unique numbers. If x is chosen randomly
[#permalink]
09 Sep 2013, 00:39