Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If 2 DIFFERENT numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 2 numbers selected is an even number?

A. 1/2
B. 1/3
C. 2/3
D. 1/4
E. 3/4

\[{\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{
\,{\rm{first}} = 2 = {\rm{even}} \hfill \cr
\,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,\]
\[? = P\left( {2\,\,{\text{different}}\,\,{\text{selected}}\,\,{\text{have}}\,{\text{sum}}\,\,{\text{even}}} \right) = P\left( {{\text{number}}\,\,{\text{2}}\,\,{\text{is}}\,\,{\text{not }}\,{\text{selected}}} \right)\]
\[{\text{total}} = C\left( {8,2} \right)\,\,\,{\text{equiprobable}}\]
\[{\text{favorable}}\,{\text{ = }}\,{\text{C}}\left( {7,2} \right)\,\,\,\,\,\,\left[ {{\text{number}}\,{\text{2}}\,\,{\text{is}}\,{\text{not}}\,{\text{an}}\,{\text{option}}} \right]\,\,\,\]
\[? = \frac{{C\left( {7,2} \right)}}{{C\left( {8,2} \right)}} = \frac{{7 \cdot 6}}{{8 \cdot 7}} = \frac{3}{4}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>

In order for the sum to be even, both primes selected must be odd. As 2 is the only even prime number, the number of selections with an even sum is equal to the number of ways to select 2 numbers from these 7 odd prime numbers, or 7C2.
The total number of selections of 2 prime numbers from the first 8 prime numbers is 8C2.
Therefore, the probability that the sum of the two numbers selected is even is
7C2 / 8C2 = {(7*6)/(1*2)}/{(8*7)/(1*2)} = 6/8 = 3/4.

Therefore, the answer is E.
Answer: E