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When 2 numbers are selected from the integers from 1 to 21 inclusive,

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Math Revolution GMAT Instructor
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When 2 numbers are selected from the integers from 1 to 21 inclusive,  [#permalink]

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New post 31 Jul 2018, 00:43
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[Math Revolution GMAT math practice question]

When \(2\) numbers are selected from the integers from \(1\) to \(21\) inclusive, what is the probability that the \(2\) selected numbers are prime numbers?

\(A. \frac{2}{15}\)
\(B. \frac{1}{5}\)
\(C. \frac{1}{3}\)
\(D. \frac{4}{15}\)
\(E. \frac{2}{5}\)

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Re: When 2 numbers are selected from the integers from 1 to 21 inclusive,  [#permalink]

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New post 31 Jul 2018, 01:00
MathRevolution wrote:
[Math Revolution GMAT math practice question]

When \(2\) numbers are selected from the integers from \(1\) to \(21\) inclusive, what is the probability that the \(2\) selected numbers are prime numbers?

\(A. \frac{2}{15}\)
\(B. \frac{1}{5}\)
\(C. \frac{1}{3}\)
\(D. \frac{4}{15}\)
\(E. \frac{2}{5}\)



Prime numbers till 21 - 2, 3, 5, 7, 11, 13 17 19 so 8 of them..
Total ways to select 2 out of 21..=21*20

Probability=8*7/21*20=2/15

A
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Re: When 2 numbers are selected from the integers from 1 to 21 inclusive,  [#permalink]

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New post 31 Jul 2018, 01:03
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

When \(2\) numbers are selected from the integers from \(1\) to \(21\) inclusive, what is the probability that the \(2\) selected numbers are prime numbers?

\(A. \frac{2}{15}\)
\(B. \frac{1}{5}\)
\(C. \frac{1}{3}\)
\(D. \frac{4}{15}\)
\(E. \frac{2}{5}\)



1 is not a prime number

prime numbers between 1 and 21(inclusive) = 2,3,5,7,11,13,17,19=8
2 out of 8 can be selected in 8c2 ways = 28
2 out of 21 can be selected in 21c2 ways =210
probability of selecting 2 primes = 28/210=2/15
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Re: When 2 numbers are selected from the integers from 1 to 21 inclusive,  [#permalink]

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New post 02 Aug 2018, 02:20
=>

There are \(8\) prime numbers between \(1\) and \(21\), inclusive: \(2, 3, 5, 7, 11, 13, 17\) and \(19\).
So, there are 8C2 ways of selecting \(2\) numbers from these \(8\) prime numbers.
There are 21C2 ways of selecting \(2\) numbers from the \(21\) numbers from \(1\) to \(21\), inclusive.
Thus, the probability that the \(2\) selected numbers are prime numbers is 8C2 / 21C2 = ( 8*7 / 1*2) / ( 21*20 / 1*2 ) = 8*7 / 21*20 = 2/15.

Therefore, the answer is A.

Answer: A
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Re: When 2 numbers are selected from the integers from 1 to 21 inclusive, &nbs [#permalink] 02 Aug 2018, 02:20
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