If 3 different numbers are selected from the first 8 prime numbers, wh

Since the sum is even , it must contain 2 as one of the numbers. Remaining numbers can be selected in 7c2 ways
Probability = 7c2/8c3 = 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 = \(\frac{3}{8}\).

\({\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{\\
\,{\rm{first}} = 2 = {\rm{even}} \hfill \cr \\
\,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,\)

\(? = {\mathop{\rm P}\nolimits} \left( {{\rm{3}}\,\,\underline {{\rm{different}}} \,\,{\rm{selected}}\,\,{\rm{have}}\,\,{\rm{sum}}\,\,{\rm{even}}} \right) = {\mathop{\rm P}\nolimits} \left( {{\rm{number}}\,{\rm{2}}\,\,{\rm{is}}\,\,{\rm{selected}},\,\,{\rm{other}}\,\,2\,\,{\rm{free}}} \right)\)


\({\rm{total}} = C\left( {8,3} \right)\,\,\,{\rm{equiprobable}}\)

\({\rm{favorable}}\,\,{\rm{ = }}\,\,{\rm{C}}\left( {7,2} \right)\,\,\,\,\left[ {2\,\,{\rm{had}}\,\,{\rm{already}}\,\,{\rm{stolen}}\,\,{\rm{a}}\,{\rm{place,}}\,\,{\rm{have}}\,\,{\rm{7}}\,\,{\rm{options}}\,\,{\rm{for}}\,\,{\rm{the}}\,{\rm{ other}}\,\,{\rm{two }}\,{\rm{numbers}}\,} \right]\)

\(? = {{{{7!} \over {2!\,\,5!}}} \over {{{8!} \over {3!\,\,5!}}}} = {{7!\,\,3!} \over {2!\,\,8!}} = {3 \over 8}\)


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

Regards,
Fabio.

[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?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)

total possibilities: 8*7*6=336
odd sum possibilities: 7*6*5=210
even sum possibilities: 336-210=126
126/336=3/8
E

We recall that odd + odd + odd = odd and that even + odd + odd = even.

The first 8 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19. There are 8C3 = (8 x 7 x 6)/(3 x 2) = 56 ways to choose 3 of them. However, in order for the sum of the 3 numbers to be even, one of them must be 2 (the other two numbers can be any two of the remaining 7 odd primes). The number of ways of choosing 2 odd primes from 7 odd primes is 7C2 = (7 x 6)/2 = 21. Therefore, the probability that the sum of the 3 numbers selected is even is 21/56 = 3/8.

Answer: E

=>

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 \(\frac{21}{56} = \frac{3}{8}.\)

Therefore, the answer is E.
Answer: E

I went with complement method.
Probability of all odd numbers being selected= 7/8*6/7*5/6=5/8--- 5/8th probability that sum of selected numbers is odd
Probability of selecting the only even prime number= 1-5/8= 3/8 the probability that sum of selected number is even

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