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If 2 numbers are selected from the first 8 prime numbers, wh

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by Max@Math Revolution » Tue Oct 09, 2018 11:50 pm

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[Math Revolution GMAT math practice question]

If 2 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
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by fskilnik@GMATH » Wed Oct 10, 2018 4:35 am
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Fri Oct 12, 2018 7:27 am
=>

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
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by Scott@TargetTestPrep » Sat Oct 13, 2018 5:14 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If 2 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
The first 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19. We see that all of them are odd numbers except 2, and, in order for the sum of the 2 numbers selected to be even, the two numbers must be odd. Therefore, the probability is:

7/8 x 6/7 = 6/8 = 3/4

Answer: E

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