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Re: M10-14 [#permalink]
Expert Reply
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]
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]
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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]
why 1 is not considered as a prime number?
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Re: M10-14 [#permalink]
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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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Hope it helps.
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Re: M10-14 [#permalink]
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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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M10-14 [#permalink]
I find 4P1*3P1 = 12 when I calculate the total times that two numbers are drawn at random one after the other, with replacement. Then, I find 3 pairs of number (2;3),(2;5),(2;7),(3;2),(5;2),(7;2) that the sum is odd 
=> probability : 6/12=1/2
What is my mistake here?
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Re: M10-14 [#permalink]
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phamminh2003 wrote:
I find 4P1*3P1 = 12 when I calculate the total times that two numbers are drawn at random one after the other, with replacement. Then, I find 3 pairs of number (2;3),(2;5),(2;7),(3;2),(5;2),(7;2) that the sum is odd 
=> probability : 6/12=1/2
What is my mistake here?

­
When picking with replacement, we can get 16 pairs, not 12. The numbers can repeat:

(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]
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