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If a, b, and c are integers randomly chosen from the set of prime numb

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If a, b, and c are integers randomly chosen from the set of prime numb  [#permalink]

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New post 03 Jan 2019, 11:16
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

86% (01:26) correct 14% (01:15) wrong based on 41 sessions

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Re: If a, b, and c are integers randomly chosen from the set of prime numb  [#permalink]

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New post 03 Jan 2019, 12:10
3
first, a probability ranges from 0 (impossible) to 1 (certain). this fact instantly eliminates C, D and E.

second, we are choosing among odd numbers, and odd*odd + odd = even. As 23 is odd, it can impossibly be an outcome.
this is a zero possibility.

the answer is 'A'.
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Re: If a, b, and c are integers randomly chosen from the set of prime numb  [#permalink]

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New post 03 Jan 2019, 12:11
Top Contributor
Bunuel wrote:
If a, b, and c are integers randomly chosen from the set of prime numbers greater than 2 and less than 30, what is the probability that ab + c is equal to 23?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Key concept: All prime numbers greater than 2 are ODD (3, 5, 7, 11, 13, 17, 19, 23, 29)

So, ab + c = (ODD)(ODD) + (ODD) = ODD + ODD = EVEN

This means that the sum ab + c can NEVER equal an odd number (like 23)

So, P(ab + c equals 23) = 0

Answer: A

Cheers,
Brent
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Re: If a, b, and c are integers randomly chosen from the set of prime numb  [#permalink]

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New post 06 Jan 2019, 04:21
Bunuel wrote:
If a, b, and c are integers randomly chosen from the set of prime numbers greater than 2 and less than 30, what is the probability that ab + c is equal to 23?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


for all values of given set ab+c would always give an even no.

options c,d,e are all wrong since probabilty can only be 1 or 0

IMO A , 0 is correct
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Re: If a, b, and c are integers randomly chosen from the set of prime numb   [#permalink] 06 Jan 2019, 04:21
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