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

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}

CounterSniper · Answer

we cannot have all 3 odd numbers 
because o+o+o =o
therefore, one even prime number , which is 2 is always required .
no of ways we can select 2 = 1
no of ways we can select the remaining =  7c2  = 21
no of ways we can select 3 =  8c3  = 56
p(sum is even) =  \frac{21}{56} =  \frac{3}{8} 
E

BrentGMATPrepNow · Answer

The first 8 prime numbers are: {2, 3, 5, 7, 11, 13, 17, 19}
Notice that only one prime number is EVEN, and the remaining seven numbers are ODD. 
Also notice that the sum of 3 numbers will be EVEN  only if  one of the three selected numbers is 2
So, the question is really asking "What is the probability that 2 is among the three selected values?"

There are many ways to answer this question. 
A super-quick approach is to  use the complement  
That is, P(Event A happening) = 1 - P(Event A  not  happening)
So, we can write: P(2 is among the three selected values) = 1 -  P(2 is NOT among the three selected values)

P(2 is NOT among the three selected values)  
P(2 is NOT among the three selected values) = P(all three selected values are ODD)
= P(1st number is odd   AND   2nd number is odd   AND   3rd number is odd)
= P(1st number is odd)   x   P(2nd number is odd)   x   P(3rd number is odd)
= 7/8   x   6/7   x   5/6
=  5/8

So, P(2 is among the three selected values) = 1 -  5/8 
= 3/8

Answer: E

Cheers,
Brent

pandeyashwin · Answer

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

GMATGuruNY · Answer

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} .

E

fskilnik · Answer

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

? = {\mathop{m P}
olimits} \left( {{m{3}}\,\,\underline {{m{different}}} \,\,{m{selected}}\,\,{m{have}}\,\,{m{sum}}\,\,{m{even}}} ight) = {\mathop{m P}
olimits} \left( {{m{number}}\,{m{2}}\,\,{m{is}}\,\,{m{selected}},\,\,{m{other}}\,\,2\,\,{m{free}}} ight)

{m{total}} = C\left( {8,3} ight)\,\,\,{m{equiprobable}}

{m{favorable}}\,\,{m{ = }}\,\,{m{C}}\left( {7,2} ight)\,\,\,\,\left

? = {{{{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.

gracie · Answer

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

ScottTargetTestPrep · Answer

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

MathRevolution · Answer

=>

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  7 C 2  = 21.
The total number of selections is  8 C 3  = 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

taniad · Answer

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

bumpbot · Answer

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

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