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Re: What is the probability of selecting a number that has exactly ... [#permalink]
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The OA MUST be posted always.

At the very least, use the feature to reveal the OA after a while, postponing the OA itself
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Re: What is the probability of selecting a number that has exactly ... [#permalink]
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The answer is C but in A we do have 1/3

Should be 1/9
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Re: What is the probability of selecting a number that has exactly ... [#permalink]
Carcass wrote:
The answer is C but in A we do have 1/3

Should be 1/9
or 3/9 reduced to 1/3, otherwise the question becomes selves-made)

Your answer is accepted as the correct one.
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Re: What is the probability of selecting a number that has exactly ... [#permalink]
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Finding the right answer to this question requires us to do 2 things :
1. Identify the numbers having exactly 3 factors
2. Finding the probability that if you pick any number randomly from the list, it belongs to the one in list #1

1. We all know that any integer is divisible by 1 and itself. That means 2 factors are bare minimum to any number. So, to identify the number with 3 factors, we just need to identify the numbers from the given list that have only one factor apart from 1 and itself.
Now, since all the numbers in the given list are perfect squares, the only integers that will have 1 extra factor in addition to 1 and itself would be the prime numbers. This is because composite numbers can be factorized into 2 or more integers, so we would never have 'only' 1 additional factor for composites.
For example,
for primes,
49 = 7^2 ; factors = {1,7,49}
25 = 5^2 ; factors = {1,5,25}

and for composites,
36 = 6^2 ; factors = {1,2,4,3,9,12,18,36}

2. since we have 4 numbers out of total 9 numbers as the squares of prime numbers ( 5^2 ,7^2 , 11^2 ,13^2 ) our probability would be 4/9.
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Re: What is the probability of selecting a number that has exactly ... [#permalink]
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motion2020 wrote:
What is the probability of selecting a number that has exactly three factors from the following set of numbers?

{25, 36, 49, 64, 81, 100, 121, 144, 169}

(A) 1/3
(B) 2/9
(C) 4/9
(D) 5/9
(E) 7/9


A number with exactly three factors must be a perfect square of a prime number

All the numbers are perfect squares, but we need to look for squares of ONLY prime numbers
i.e. \(5^2, 7^2, 11^2\), and \(13^2\)

So, the required probability = \(\frac{4}{9}\)

Hence, option C
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Re: What is the probability of selecting a number that has exactly ... [#permalink]
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the only positive integers with exactly three factors are the squares of primes. For instance, the factors of 9 are 1, 3, and 9, and the factors of 49 are 1, 7, and 49.
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Re: What is the probability of selecting a number that has exactly ... [#permalink]
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