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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]

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01 Sep 2006, 00:40
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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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01 Sep 2006, 02:03
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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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01 Sep 2006, 02:48
C for me.

P(product < 31) =1/8C3

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

Diff = 34/56 or 17/28
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01 Sep 2006, 04:49
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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01 Sep 2006, 04:50
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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 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]

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19 Sep 2013, 01:59
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31 Oct 2013, 02:08
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?

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25 Nov 2013, 12:55
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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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07 May 2014, 02:34
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: Set S is the set of all prime integers between 0 and 20. If [#permalink]

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01 Jun 2015, 22:28
Hello from the GMAT Club BumpBot!

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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]

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23 Aug 2015, 23:48
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S = 2,3,5,7,11,13,17,19
Size of S = 8

first number can be chosen in 8 ways
2nd number in 7 ways
3rd number 6 ways

so total outcomes = 8 x 7 x 6 = 336 possible triplets

for the sum to be odd all three must be odd = 7 x 6 x 5 = 210 ways

prob of odd = 210/336 = 5/8

prod less than 31 can only be possible if we have 2,3,5 this can be done in 3x2x1 ways = 6 ways
prob (sum<31) = 6/8x7x6 = 1/56

diff = 5/8 - 1/56 = 34/56 = 17/28
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Set S is the set of all prime integers between 0 and 20. If [#permalink]

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24 Aug 2015, 01:47
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I did it this way:

Primes between 0 and 20:
2, 3, 5, 7, 11, 13, 17, 19

I: Probability that the product of (x,yz) < 31:
Only possible way is 2*3*5 = 30, all other products are >31

=> $$1/8 * 1/7 * 1/6 * 3! = 3*2*1/336 = 1/56$$

II: Probability that the sum of (x+y+z) = odd:
There are numerous possibilities as long as 2 (the only even number) is not included.

=> $$7/8 * 6/7 * 5/6 = 5/8$$

III: I - II:

=> $$1/56 - 5/8 = |-17/28|$$
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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]

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31 Aug 2015, 22:23
waltiebikkiebal 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

Im not understanding how you come to 35/56?

We have a total of 8 prime nos between 1 and 20. To have an odd sum, the three nos should be odd. Out of the total 8 prime nos, there are 7 odd nos, so the no of ways to select 3 nos out of 7 would be 7C3.
Hence, the probability would be 7C3/8C3 = 35/56
Re: Set S is the set of all prime integers between 0 and 20. If   [#permalink] 31 Aug 2015, 22:23
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