*Here’s this month’s GRE Brain Twister! Try to answer it in the comments below, and we’ll post the answer on Thursday!*

A math teacher assigns a distinct prime number, starting with the lowest, for each student in her class. If she chooses two students at random, the probability the sum of their numbers is not even is less than 1/10. What is the fewest possible number of students in her class?

(A) 19

(B) 20

(C) 21

(D) 71

(E) 73

### Most Popular Resources

As two is the only even prime number, and to yield an odd number we must add an even to an odd, the only way the sum of the two chosen kids’ numbers will turn out odd is if one of these kids’ numbers is two. This kid is a singular entity, yet there are two possibilities the kid will be chosen, therefore to yield a 1/10 probability that the kid will be chosen, we must multiply the fraction by 2, yielding 2/20 — 2 possibilities the kid will be chosen over 20 total kids. So, to yield exactly a 1/10 probability there must be 20 kids in the class. But the quesiton asks for less than 1/10. To yield less than a 1/10 probability, there must be more than 20 kids in the class. Therefore, the answer is (C) 21.

Answer is (C) 21.

I think the answer is C) 21

if # of students = 21 then,

probability = (1 – (20/21)(19/20)) = (1 – 0.904) which less than 1/10, so far this answer is viable but in order to make sure this is the fewet possible # of students I will check choice B

– using 20 students, the probability = (1 – (19/20)(18/19)) which equals exactly 1/10, hence choice B is incorrect

– checking choice D and E is not needed, because we are looking for the fewest number of students. and calculating the probability using choice A results in a number larger than 1/10

21?

I believe the answer is C. In order for the sum of the numbers to be not even we must have a two (because it is the only even prime) and another odd prime number. From there I went through the answer choices starting at A.

For of A I got (1)(18)/19C2= .105

For B I did (1)(19)/20C2=.1.

For C I got (1)(20)/21C2=.095 so my answer was C

answer n>20.. 21 students 🙂

Only way we can have the sum as odd is when 2 is added to another prime number..

so total number of ways of selecting 2 students = nC2 = (n*(n-1)/2)

no. of ways of selecting 2 and another prime number ( n-1) possibilities..

given, Probabality < (1/10)

(n-1)/(n*(n-1)/2) < (1/10)

2/n 20.. so answer should be 21 🙂