Bunuel 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. \(\frac{4}{5}\)
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.