Problem Solving

[Math Revolution GMAT math practice question]

If 3 different numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 3 numbers selected is even?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/8

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

If 3 different numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 3 numbers selected is even?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/8

First 8 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19.
The sum will be even if 2 is among the three numbers selected, with the result that the sum = EVEN PRIME + ODD PRIME + ODD PRIME = EVEN.
Since 3 NUMBERS are selected from 8 OPTIONS, the probability that 2 will be among the three numbers selected = 3/8.

The correct answer is E.

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

If 3 different numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 3 numbers selected is even?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/8

$${\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{
\,{\rm{first}} = 2 = {\rm{even}} \hfill \cr
\,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,$$$$? = \operatorname{P} \left( {{\text{3}}\,\,\underline {{\text{different}}} \,\,{\text{selected}}\,\,{\text{have}}\,\,{\text{sum}}\,\,{\text{even}}} \right) = \operatorname{P} \left( {{\text{number}}\,{\text{2}}\,\,{\text{is}}\,\,{\text{selected}},\,\,{\text{other}}\,\,2\,\,{\text{free}}} \right)$$
$${\text{total}} = C\left( {8,3} \right)\,\,\,{\text{equiprobable}}$$
$${\rm{favorable}}\,\,{\rm{ = }}\,\,{\rm{C}}\left( {7,2} \right)\,\,\,\,\left[ {2\,\,{\rm{had}}\,\,{\rm{already}}\,\,{\rm{stolen}}\,\,{\rm{a}}\,{\rm{place,}}\,\,{\rm{have}}\,\,{\rm{7}}\,\,{\rm{options}}\,\,{\rm{for}}\,\,{\rm{the}}\,{\rm{ other}}\,\,{\rm{two }}\,{\rm{numbers}}\,} \right]$$
$$? = \frac{{\frac{{7!}}{{2!\,\,5!}}}}{{\frac{{8!}}{{3!\,\,5!}}}} = \frac{{7!\,\,3!}}{{2!\,\,8!}} = \frac{3}{8}$$

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

Regards,
Fabio.

=>

Suppose p, q and r are prime numbers.
In order for p + q + r to be even, one of them must equal 2, since 2 is the only even prime number.
Once 2 has been selected, there are 7 prime numbers remaining from which to select 2 numbers. Thus, the number of selections in which the sum of the 3 numbers is even is 7C2 = 21.
The total number of selections is 8C3 = 56.
Thus, the probability that the sum of the three numbers is even is 21/56 = 3/8.

Therefore, the answer is E.
Answer: E