Bunuel wrote:
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
Let the assignment be for 2 students then 2, 3 probability of even sum (odd+odd is required) is 0 (2 is compulsory, rest all prime numbers are odd)
Let the assignment be for 3 students then 2, 3, 5 probability of even sum is 1/3
Let the assignment be for 4 students then 2,3,5,7 probability of even sum is 1/2
Let the assignment be for 5 students then 2,3,5,7,11 probability of even sum is 6/10
Now we know
(n-1)C2/nC2 = Probability of sum even
for n = 19
18C2/19C2 = 9*17/19*9 = 17/19 not even will be 2/19
for n=20
19*9/19*10 = 9/10 not even will be 1/10
for n=21
10*19/21*10 = 19/21 not even will be 2/21
for n=71
35*69/35*71 = 69/71 not even will be 2/71
for n = 73
72C2/73C2 = 71/73 not even will be 2/73
Fewest will be 21
Option C
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