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

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C for me.

P(product < 31) =1/8C3

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

Diff = 34/56 or 17/28
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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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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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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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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
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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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Set S is the set of all prime integers between 0 and 20. If [#permalink]
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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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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
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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]
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Hello Friends,
I am really confused by this question and here's why: it states "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"

So shouldn't this mean that the triplet should satisfy BOTH conditions?
I thought like this and did following:
(i) For product of a*b*c < 31 we need 2 BUT
(ii) for sum to be odd ie a+b+c = odd we need 3 odd numbers.
Thus, P(a*b*c < 31) = 0 and P(a+b+c = o) = (7C3)/(8C3) = 35/56 = 5/8.
Clearly, this doesn't seem to be correct according to answer choices. So my question to you'll is:

q) how did you arrive at the conclusion that triplets for both conditions shouldn't be the same

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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]
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manishtank1988 wrote:
Hello Friends,
I am really confused by this question and here's why: it states "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"

So shouldn't this mean that the triplet should satisfy BOTH conditions?
I thought like this and did following:
(i) For product of a*b*c < 31 we need 2 BUT
(ii) for sum to be odd ie a+b+c = odd we need 3 odd numbers.
Thus, P(a*b*c < 31) = 0 and P(a+b+c = o) = (7C3)/(8C3) = 35/56 = 5/8.
Clearly, this doesn't seem to be correct according to answer choices. So my question to you'll is:

q) how did you arrive at the conclusion that triplets for both conditions shouldn't be the same

Hi there,

Don't worry about "the product less than 31" part. Just focus on what the question asks: The probability that the sum will be odd.
In order for the sum to be odd, you need either 2 evens and one odd, or 3 odds. Since all the prime numbers only have one even number that is 2, 2 evens and one odd is not an option. So just focus on 3 odds.
There are only 7 odds prime number between 0 and 20. Hence, the answer is 7C3/8C3.
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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]
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manishtank1988 wrote:
Hello Friends,
I am really confused by this question and here's why: it states "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"

So shouldn't this mean that the triplet should satisfy BOTH conditions?
I thought like this and did following:
(i) For product of a*b*c < 31 we need 2 BUT
(ii) for sum to be odd ie a+b+c = odd we need 3 odd numbers.
Thus, P(a*b*c < 31) = 0 and P(a+b+c = o) = (7C3)/(8C3) = 35/56 = 5/8.
Clearly, this doesn't seem to be correct according to answer choices. So my question to you'll is:

q) how did you arrive at the conclusion that triplets for both conditions shouldn't be the same

Hi again,

I apologize -- I just noticed that part of what I had written is missing.

So I was saying that the probability that the sum of the 3 numbers is odd is 7C3/8C3.
Now let's focus on the remaining of the question: the positive difference between the above probability - 7C3/8C3 - and the probability that the product of these 3 numbers is a number less than 31.
There is only one set of numbers from the list that gives a product less than 31. That set of number is 2,3,5 - 2*3*5=30. Any other set would give a larger number. That being said, substract 3C3/8C3 from 7C3/8C3 and you get the answer.

Hope it helps.

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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]
If set S is the set of all prime integers between 0 and 20 then:
S = {2, 3, 5, 7, 11, 13, 17, 19}
Let’s start by finding the probability that the product of the three numbers chosenis a number less than 31. To keep the product less than 31, the three numbers must be 2, 3 and 5. So, what is the probability that the three numbers chosen will be some combination of 2, 3, and 5?
Here’s the list all possible combinations of 2, 3, and 5:
case A: 2, 3, 5
case B: 2, 5, 3
case C: 3, 2, 5
case D: 3, 5, 2
case E: 5, 2, 3
case F: 5, 3, 2
This makes it easy to see that when 2 is chosen first, there are two possible combinations. The same is true when 3 and 5 are chosen first. The probability of drawing a 2, AND a 3, AND a 5 in case A is calculated as follows (remember, when calculating probabilities, AND means multiply):
case A: (1/8) x (1/7) x (1/6) = 1/336
The same holds for the rest of the cases.
case B: (1/8) x (1/7) x (1/6) = 1/336
case C: (1/8) x (1/7) x (1/6) = 1/336
case D: (1/8) x (1/7) x (1/6) = 1/336
case E: (1/8) x (1/7) x (1/6) = 1/336
case F: (1/8) x (1/7) x (1/6) = 1/336
So, a 2, 3, and 5 could be chosen according to case A, OR case B, OR, case C, etc. The total probability of getting a 2, 3, and 5, in any order, can be calculated as follows (remember, when calculating probabilities, OR means add):
(1/336) + (1/336) + (1/336) + (1/336) + (1/336) + (1/336) = 6/336
Now, let’s calculate the probability that the sum of the three numbers is odd. In order to get an odd sum in this case, 2 must NOT be one of the numbers chosen. Using the rules of odds and evens, we can see that having a 2 would give the following scenario:
even + odd + odd = even
So, what is the probability that the three numbers chosen are all odd? We would need an odd AND another odd, AND another odd:
(7/8) x (6/7) x (5/6) = 210/336
The positive difference between the two probabilities is:
(210/336) – (6/336) = (204/336) = 17/28
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Set S is the set of all prime integers between 0 and 20. If [#permalink]
ends up being:

3C3/8C3 - 7C3/8C3 = 1/56 - 5/56 = -17/24 --> 17/28

(1) probability product less than 31
only combo works with 2, 3, 5 --> 3C3
Total combinations 8C3
Probability = 3C3/8C3 = 1/56

(2) probability odd
Any combo that doesn't have 2
7C3 ways to pick without choosing 2
8C3 total combos
Probability = 7C3/8C3 = 5/8

5/8-1/56 = 17/28
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Re: Set S is the set of all prime integers between 0 and 20. If [#permalink]
@GMATT73Given: Set S is the set of all prime integers between 0 and 20.
Asked: 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?
S = {2,3,5,7,11,13,17,19}
Number of ways to chose 3 numbers out of 8 = 8C3 = 56

Products less than 31 = {2*3*5=30} ; All other products are greater than 31
The probability that the product of these three numbers is a number less than 31 = 1/56

For the sum of these three numbers is odd
When all three numbers are odd; Number of ways such that all three numbers are odd =  7C3 = 35
Number of ways such that 2 numbers are even and one is odd = 0
The probability that the sum of these three numbers is odd = 35/56

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 = |1/56 - 35/56| = 34/56 = 17/28

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

I actually thought the three numbers that give the product less than 31 should also give the product as odd.
So for case the prob is 1/56 and for the second prob is 0. (as we have 2, the sum cant be odd)

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