M15-10

Question

Set  S  consists of all prime integers less than 10. If two different numbers are chosen from set  S  at random, what is the probability that the product of these numbers will be greater than the product of the numbers that were not chosen?

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

Bunuel · Accepted Answer

Official Solution:

Set  S  consists of all prime integers less than 10. If two different numbers are chosen from set  S  at random, what is the probability that the product of these numbers will be greater than the product of the numbers that were not chosen?

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

Set  S  is defined as  \{2, 3, 5, 7\} .
 
The most straightforward way to address this question is to recognize that we are selecting half of the numbers from the set (essentially dividing the group of four into two smaller groups of two). As a tie is not possible, the probability that the product of the numbers in one subgroup exceeds that in the other subgroup is  \frac{1}{2} , because neither subgroup is favored over the other in terms of probability.

Answer: C

sidomania · Answer

Prime numbers less than 10: 2,3,5,7==>4 numbers in set  S

2 are selceted==> 4!/2!*2!= 6 ways of selecting two numbers from set  S

Favourable outcomes: 3-5,3-7,5-7==>3

P=3/6 =1/2  Ans. C

filipTGIM · Answer

Set : 2,3,5,7.

Recognise that all combinations involving 3,5,7 form a product greater then the other numbers not chosen.

Probability of picking 3,5 or 7 as the first number : 3/4
Probability of picking 3,5 or 7 as the second number : (3-1)/(4-1) = 2/3

3/4 * 2/3 = 1/2

naumyuk · Answer

To solve this problem it's important to understand that there cannot be a situation in which product of two chosen integers is equal to product of two left integers. It's can be concluded from the fact that all integers in the set are prime integers.

There can be 6 different products, since 4C2 = 6. And in 3 cases product of chosen integers will be greater than product of left integers and in 3 cases product of chosen integers will be less than product of left integers.

Therefore the probability is exactly 1/2.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

LearningCurve · Answer

I just went with the straightforward approach of making the 6 cases. 2 will go with 3 5 7 ; 3 will go with 5 and 7 ; and 5 will how with 7. 
This is the approach of selecting cases wherein I have covered all possible cases . 3 of 6 cases give us an integer greater than that formed by prime integers not chosen. So 1/2. Took me less than a minute on paper, so didn’t need p&c or any probability formula.

Posted from my mobile device

sssdd1234 · Answer

I think this is a  high-quality  question and the explanation isn't clear enough, please elaborate. Could you expand a bit more on how we can get 1/2 without trying out the different combos? I did not understand the explanation given. Thank you so much for the help!

Bunuel · Answer

We are dividing the group of four numbers into two equal subgroups of two. Since a tie isn’t possible, one of the groups will always have the greater product. Each subgroup has an equal chance of having the greater product. Therefore, the probability that the product of the numbers in one subgroup exceeds that of the other is 1/2, as neither subgroup has any inherent advantage. You can explore more conventional methods above or in this thread: https://gmatclub.com/forum/set-s-consis ... 24713.html Hope it helps.

sssdd1234 · Answer

Thank you so much!

IZ19 · Answer

Why 1 is not included in set S? Set S is made up of all prime integers lower than 10 and 1 fulfills all the criteria

Bunuel · Answer

1 is not a prime number. The smallest prime is 2.

CarlRH · Answer

One way to think about it, at least for me is:

Imagine in round one you divided the group into 2 subgroups (ex (2,3) and (5,7)). Now, if you pick one of those two sub groups, your chances of winning is 50%. 
In other words, if you choose subgroup one you will lose and in round two if you pick subgroup two, you will win. So you win one out of two trials hence you have a winning of 50%. And winning is equivalent to the pair that has the greatest product.

Now we know that any other two sub groups will have the same results. Therefore, the probability does not change and your probability remains at 50%.

I hope this helps

Fisayofalana · Answer

I like the solution - it’s helpful. Old-fashioned way: (2,3) (3,2) (2,5) (5,2) (2,7) (5,7) (3,5) (5,3) (3,7) (7,3) (7,5) (7,2). Total outcomes = 12. Total favorable outcomes = 6. 6/12 = 1/2.

M15-10

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