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Math Expert V
Joined: 02 Sep 2009
Posts: 55150
A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 43% (02:31) correct 57% (02:49) wrong based on 39 sessions

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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

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Manager  B
Joined: 02 Nov 2015
Posts: 163
GMAT 1: 640 Q49 V29 Re: A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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2
The first 23 prime numbers 2,3,5,7,9,....67, 71 and 73. Selecting 2 out of 23is 23C2 means 253
And except 2 if we add any number to the other we will stand ven number only. So total outcomes are (2,3), (2,5), (2,7)......(2,73), leading to total of 20 outcomes.

So probability is 20/253 which is less than 1/10.

But if we select 19 or 20 numbers then the probability turns out to be ≥1/10 which is not acceptable as per te question stem.
So my Answer is C.

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Current Student P
Joined: 18 Aug 2016
Posts: 618
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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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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Luckisnoexcuse
Math Expert V
Joined: 02 Aug 2009
Posts: 7667
Re: A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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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

Hi..
The proper method would be
let n be the number, so n-1 are ODD prime..
1) choosing any 2 out of n is nC2..
2) ways in which sum is not even..
one has to be 2 and other any from n-1.. so n-1C1..

Probability =$$\frac{n-1C1}{nC2}= \frac{(n-1)!}{(n-1-1)!1!}*\frac{(n-2)!2!}{n!}=(n-1)*\frac{2}{n(n-1)}=\frac{2}{n}$$..

so from Q $$\frac{2}{n}<\frac{1}{10}$$ or n>20..
so ans is n=21
C
_________________ Re: A math teacher assigns a distinct prime number, starting with the lowe   [#permalink] 27 Jul 2017, 07:31
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# A math teacher assigns a distinct prime number, starting with the lowe

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