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m15,#10

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m15,#10 [#permalink] New post 28 Nov 2008, 01:10
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Set S consists of all prime integers less than 10. If two numbers are chosen form set S at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen?

(A) \frac{1}{3}
(B) \frac{2}{3}
(C) \frac{1}{2}
(D) \frac{7}{10}
(E) \frac{4}{5}

[Reveal] Spoiler: OA
C

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What is wrong with my approach?
The set consists of (2,3,5,7)
Niow acc to soncition the nos satisfying are (5,7), (3,5) and (3,7)
Thus the probabiliy is (1/4*1/4)+(1/4*1/4)+(1/4*1/4)=3/16

But this is not the ans :(
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Re: m15,#10 [#permalink] New post 28 Nov 2008, 02:21
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It should be (1/4*1/3)+(1/4*1/3)+(1/4*1/3) = 3/12 = 1/4.

But, the sequence is immaterial here as we are interested in sum. Hence, (5,3) and (3,5) will give the same result. Thus, multiply 1/4 by 2 and 1/2 should be the answer.

Alternatively, I will solve it as follows:

favorable outcome = 3 (as you have listed earlier).

Since, order does not matter, total outcome = 4C2 = 6

Hence, probability = 3/6 = 1/2
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Re: m15,#10 [#permalink] New post 28 Nov 2008, 02:31
I would go with exactly the same approach as the second one stated by scthakur.

Total outcomes = 2 numbers out of 4 hence 4C2 =6

Possible outcomes = 2 numbers out of 3 hence 3C2 = 3

Probability = 3/6 = 1/2

Whats the QA ?
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Re: m15,#10 [#permalink] New post 28 Nov 2008, 04:16
scthakur, nice explanation

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Re: m15,#10 [#permalink] New post 07 Dec 2008, 19:14
We now the total possibilities is 4c2

List out the possibilities

Chossen Not choosen
23 57 - no
25 37 -no
27 35 - no

35 27 -yes
37 25 - yes

57 23 - yes

4c2=6

3/6 = 1/2
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Re: m15,#10 [#permalink] New post 21 Dec 2009, 06:37
The total number of combinations is the following: 4!/(2!*2!) = 6

The only ones that work are these: 3*5, 3*7, 5*7

3/6 = 1/2
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Re: m15,#10 [#permalink] New post 21 Dec 2009, 06:59
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All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2.

3/4 * 2/3 = 6/12 = 1/2

So C.
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Re: m15,#10 [#permalink] New post 21 Dec 2009, 19:29
Ans : C

there are only two possibal out come and have 50% chance of geting either one
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Re: m15,#10 [#permalink] New post 28 Dec 2010, 09:56
C for me as well. :)
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Re: m15,#10 [#permalink] New post 28 Dec 2010, 20:36
i think the ans is 1/3

Pls check whether my understanding is correct

We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)

There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.

So p= 2/4C2 = 2/6 = 1/3 Ans
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Re: m15,#10 [#permalink] New post 28 Dec 2010, 23:19
GUD eXPLANATION..
its C.
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Re: m15,#10 [#permalink] New post 29 Dec 2010, 23:27
C. The limitations of the set allow you to list the specific product options and compare them. I'm someone who's skilling up in quant and since I know that I make simple mistakes, I opted to actually write out the product options.
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Re: m15,#10 [#permalink] New post 30 Dec 2010, 01:12
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ritula wrote:
Set S consists of all prime integers less than 10. If two numbers are chosen form set S at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen?

(A) \frac{1}{3}
(B) \frac{2}{3}
(C) \frac{1}{2}
(D) \frac{7}{10}
(E) \frac{4}{5}

[Reveal] Spoiler: OA
C

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S={2, 3, 5, 7}

The simplest way would be to realize that we choose half of the numbers (basically we divide the group of 4 into two smaller groups of 2) and since a tie is not possible then the probability that the product of the numbers in either of subgroup is more than that of in another is 1/2 (the probability doesn't favor any of two subgroups).

Answer: C.
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Re: m15,#10 [#permalink] New post 03 Jan 2011, 12:17
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zahuruddin wrote:
i think the ans is 1/3

Pls check whether my understanding is correct

We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)

There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.

So p= 2/4C2 = 2/6 = 1/3 Ans


Hi!

1 isn't a prime number, which is the first reason that you're deriving a different answer.

The second reason is that you've misinterpreted the question - we're not comparing the product of the two numbers we select to the product of the non-primes less than 10; we're comparing the product of the two numbers we select to the product of the other primes less than 10.

So, our set is {2, 3, 5, 7} and we're comparing the products of two pairs of numbers from within the set.

The quickest ways to solve are to either use Bunuel's intuitive approach (we're comparing 2 vs 2 with no ties, so symmetry will lead to half the pairs being greater than the other half) or via brute force. Since there are only 6 possibilities, it doesn't take long to count them out:

2*3 vs 5*7 - not greater
2*5 vs 3*7 - not greater
2*7 vs 3*5 - not greater
3*5 vs 2*7 - greater
3*7 vs 2*5 - greater
5*7 vs 3*5 - greater

Probability = (# of desired outcomes)/(total # of possibilities) = 3/6 = 1/2
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Re: m15,#10 [#permalink] New post 30 Dec 2011, 06:06
Counting 1 as a prime number is exactly the mistake I made... Funnily enough, they have an answer choice that corresponds to this particular way of "solving" the question.
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Re: m15,#10 [#permalink] New post 30 Dec 2011, 06:14
C

total ways of choosing 2 out of 4 = !4/(!2*!2) i.e. 6
total ways the product can be greater than the other two = (3,5), (5,7), (7,3) = 3
answer=3/6 = 1/2
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Re: m15,#10 [#permalink] New post 30 Dec 2011, 07:30
I solved by calculating how many solutions do not involve 2.

1 - 1/4 * 1 - 1/3 = 6 /12 = 1/2. answer is C
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Re: m15,#10 [#permalink] New post 30 Dec 2011, 18:05
C.
The Kaplan Instructor explained it pretty well.
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Re: m15,#10 [#permalink] New post 31 Dec 2011, 04:27
Moss wrote:
All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2.

3/4 * 2/3 = 6/12 = 1/2

So C.


This one is fast and least error prone for this problem.

Thanks.
Re: m15,#10   [#permalink] 31 Dec 2011, 04:27
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