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A math teacher assigns a distinct prime number, starting with the lowe

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A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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New post 27 Jul 2017, 01:30
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Question Stats:

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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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Re: A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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New post 27 Jul 2017, 03:18
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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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Re: A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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New post 27 Jul 2017, 05:56
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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Re: A math teacher assigns a distinct prime number, starting with the lowe  [#permalink]

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New post 27 Jul 2017, 07:31
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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
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Re: A math teacher assigns a distinct prime number, starting with the lowe   [#permalink] 27 Jul 2017, 07:31
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