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Set S is the set of all prime integers between 0 and 20. If

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Set S is the set of all prime integers between 0 and 20. If [#permalink] New post 31 Aug 2006, 23:40
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Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

(A) 1/336
(B) 1/2
(C) 17/28
(D) 3/4
(E) 301/336
[Reveal] Spoiler: OA
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 [#permalink] New post 01 Sep 2006, 01:03
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The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28
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 [#permalink] New post 01 Sep 2006, 01:48
C for me.

P(product < 31) =1/8C3

P(sum is odd) = 7C3/8C3

Diff = 34/56 or 17/28
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Re: PS Sets [#permalink] New post 01 Sep 2006, 03:49
GMATT73 wrote:
Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

(A) 1/336
(B) 1/2
(C) 17/28
(D) 3/4
(E) 301/336


Ans C The only time the product will be less than 31 is when 2,3,5

1/(8C3)

The only time an even sum would occur is when 2 is included in the mix

so excluding 2 7C3/8C3

The difference will be C
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 [#permalink] New post 01 Sep 2006, 03:50
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


Scintillating :cool Don't you just love it when we can reinforce multiple concepts in one problem?

1. Rule of primes
2. Adding odd and even integers
3. Triplets
4. Dependent probabilty
5. Combinatorics
6. Positive sum (absolute value)
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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink] New post 19 Sep 2013, 00:59
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Re: [#permalink] New post 31 Oct 2013, 01:08
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


Im not understanding how you come to 35/56?

Can someone please explain?
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Re: [#permalink] New post 25 Nov 2013, 11:55
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


You got the answer but your procedure is flawed. You must INCLUDE 2 not exclude since you want the sum to be odd. Anyways you get the same combinatorics fraction.

Cheers
J :)
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Re: Re: [#permalink] New post 07 May 2014, 01:34
jlgdr wrote:
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


You got the answer but your procedure is flawed. You must INCLUDE 2 not exclude since you want the sum to be odd. Anyways you get the same combinatorics fraction.

Cheers
J :)


2+3+5= 10...?
3 numbers here..so all 3 odd is necessary

FYI if we use 2 here..i.e 2 is a necessary filler..then the no of ways you can get an even sum is 21..and probability is 7C2/8C3=21/56
Correct me if I am wrong
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Re: Re:   [#permalink] 07 May 2014, 01:34
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