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M10-14

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Bunuel
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M10-14 [#permalink] New post   16 Sep 2014, 00:42
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70% (02:08) correct 30% (02:34) wrong based on 132 sessions
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Set \(S\) consists of all prime integers less than 10. If two numbers are drawn at random one after the other, with replacement, from this set, what is the probability that the sum of these two numbers is odd?

A. \(\frac{1}{8}\)
B. \(\frac{1}{6}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{5}{8}\)
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C
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Bunuel
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Re M10-14 [#permalink] New post   16 Sep 2014, 00:42
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Set \(S\) consists of all prime integers less than 10. If two numbers are drawn at random one after the other, with replacement, from this set, what is the probability that the sum of these two numbers is odd?

A. \(\frac{1}{8}\)
B. \(\frac{1}{6}\)
C. \(\frac{3}{8}\)
D. \(\frac{1}{2}\)
E. \(\frac{5}{8}\)


Set \(S =\{2, 3, 5, 7\}\). Since the sum of the numbers must be odd, then one of the numbers must even prime, so 2, and another must be any odd prime.

\(P(\text{sum is odd}) =\)

\(=P(\text{first number is 2, another is any odd prime}) + P(\text{first number is any odd prime, another is 2})= \)

\(= \frac{1}{4}*\frac{3}{4} + \frac{3}{4}*\frac{1}{4} = \frac{6}{16} = \frac{3}{8}\).


Answer: C
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rsamant
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Re: M10-14 [#permalink] New post   05 Dec 2014, 21:54
I quickly listed out the set of numbers (2,2) (2,3) (2,5) (2,7) (3,3) (3,5) (3,7) (5,5) (5,7) and (7,7) and then I counted the only odd pair which left me with a probability of 3/10. Why is this method incorrect?
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Bunuel
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Re: M10-14 [#permalink] New post   06 Dec 2014, 06:24
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rsamant wrote:
I quickly listed out the set of numbers (2,2) (2,3) (2,5) (2,7) (3,3) (3,5) (3,7) (5,5) (5,7) and (7,7) and then I counted the only odd pair which left me with a probability of 3/10. Why is this method incorrect?


There are more possibilities when picking two numbers:

(2,2)
(2,3)
(3,2)
(2,5)
(5,2)
(2,7)
(7,2)
(3,3)
(3,5)
(5,3)
(3,7)
(7,3)
(5,5)
(5,7)
(7,5)
(7,7)

P = 6/16 = 3/8.
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Re: M10-14 [#permalink] New post   30 Aug 2018, 20:42
Hi Bunuel,

Does this 'Not necessarily different' terminology given in the question implies that the number is not repeated?
Because when I first attempted this question i got 1/2 i.e . (1/4)*(3/3) + (3/4)*(1/3) = (1/2)

But now when I saw your solution, I came to know that the numbers should not be reduced and should be kept same.
I just wanted to get it confirmed with you that everytime in future if I see this term 'Not necessarily different' should I consider that the number is not repeated?
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Re: M10-14 [#permalink] New post   30 Aug 2018, 21:00
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varmashreekanth wrote:
Hi Bunuel,

Does this 'Not necessarily different' terminology given in the question implies that the number is not repeated?
Because when I first attempted this question i got 1/2 i.e . (1/4)*(3/3) + (3/4)*(1/3) = (1/2)

But now when I saw your solution, I came to know that the numbers should not be reduced and should be kept same.
I just wanted to get it confirmed with you that everytime in future if I see this term 'Not necessarily different' should I consider that the number is not repeated?


If a number is selected from set S at random and then another number, not necessarily different, is selected from set S at random, ...

The highlighted part in the question means that from S={2,3,5,7} we can choose any number more than once. So, we can choose the following two numbers:
(2,2)
(2,3)
(3,2)
(2,5)
(5,2)
(2,7)
(7,2)
(3,3)
(3,5)
(5,3)
(3,7)
(7,3)
(5,5)
(5,7)
(7,5)
(7,7)
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Re: M10-14 [#permalink] New post   26 Sep 2022, 04:21
why 1 is not considered as a prime number?
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Re: M10-14 [#permalink] New post   26 Sep 2022, 04:53
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jayditya wrote:
why 1 is not considered as a prime number?


By definition a prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself.

The main reason 1 is not conceded as prime is because of the fundamental theorem of arithmetic (unique prime factorization theorem), which states that every integer greater than 1 can be represented uniquely as a product of prime numbers.

For example, \(60 = 2^2*3*5\) (the order of the primes does not matter here). Now, if we allow 1 to be a prime then we can represent 60 as \(1*2^2*3*5\) or as \(1^2*2^2*3*5\) or as \(1^3*2^2*3*5\) as \(1^4*2^2*3*5\) ... As you can see the representation is no longer unique and thus the fundamental theorem of arithmetic is no longer correct, which is a problem because the fundamental theorem of arithmetic is called fundamental for a reason.

2. Properties of Integers



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Re: M10-14 [#permalink] New post   05 Mar 2023, 13:41
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M10-14 [#permalink]
05 Mar 2023, 13:41
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