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# If 2 numbers are selected from the first 8 prime numbers, what is the

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If 2 numbers are selected from the first 8 prime numbers, what is the  [#permalink]

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10 Oct 2018, 01:01
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[Math Revolution GMAT math practice question]

If 2 numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 2 numbers selected is an even number?

$$A. \frac{1}{2}$$
$$B. \frac{1}{3}$$
$$C. \frac{2}{3}$$
$$D. \frac{1}{4}$$
$$E. \frac{3}{4}$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 01 Feb 2018 Posts: 30 Re: If 2 numbers are selected from the first 8 prime numbers, what is the [#permalink] ### Show Tags 10 Oct 2018, 01:43 E is the answer As We have 1st 8 prime no.s as 2,3,5,7,11,13,17 and 19 Now, we know that Odd plus odd is even So only 2 here is even which when added to any odd prime will give even sum Hence we will select 2 primes out of 7 odd primes not considering 2 So Total outcomes = 8C2 I.e 8*7 Favourable outcomes = 7C2 I.e 7*6 Hence ans is 6/8 or 3/4 Posted from my mobile device GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 Re: If 2 numbers are selected from the first 8 prime numbers, what is the [#permalink] ### Show Tags 10 Oct 2018, 05:42 1 MathRevolution wrote: [Math Revolution GMAT math practice question] If 2 DIFFERENT numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 2 numbers selected is an even number? $$A. \frac{1}{2}$$ $$B. \frac{1}{3}$$ $$C. \frac{2}{3}$$ $$D. \frac{1}{4}$$ $$E. \frac{3}{4}$$ $${\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{ \,{\rm{first}} = 2 = {\rm{even}} \hfill \cr \,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,$$ $$? = P\left( {2\,\,{\text{different}}\,\,{\text{selected}}\,\,{\text{have}}\,{\text{sum}}\,\,{\text{even}}} \right) = P\left( {{\text{number}}\,\,{\text{2}}\,\,{\text{is}}\,\,{\text{not }}\,{\text{selected}}} \right)$$ $${\text{total}} = C\left( {8,2} \right)\,\,\,{\text{equiprobable}}$$ $${\text{favorable}}\,{\text{ = }}\,{\text{C}}\left( {7,2} \right)\,\,\,\,\,\,\left[ {{\text{number}}\,{\text{2}}\,\,{\text{is}}\,{\text{not}}\,{\text{an}}\,{\text{option}}} \right]$$ $$? = \frac{{C\left( {7,2} \right)}}{{C\left( {8,2} \right)}} = \frac{{7 \cdot 6}}{{8 \cdot 7}} = \frac{3}{4}$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net VP Joined: 09 Mar 2016 Posts: 1230 Re: If 2 numbers are selected from the first 8 prime numbers, what is the [#permalink] ### Show Tags 10 Oct 2018, 07:06 1 MathRevolution wrote: [Math Revolution GMAT math practice question] If 2 numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 2 numbers selected is an even number? $$A. \frac{1}{2}$$ $$B. \frac{1}{3}$$ $$C. \frac{2}{3}$$ $$D. \frac{1}{4}$$ $$E. \frac{3}{4}$$ odd+odd = even from total 8 prime numbers, 7 numbers are odd and one is even i.e. 2 $$C^7_2$$ = 21 (choosing any two odd numbers from 7 $$C^8_2$$ = 28 ( total number of outcomes) $$\frac{21}{28}$$ i.e. $$\frac{3}{4}$$ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8011 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If 2 numbers are selected from the first 8 prime numbers, what is the [#permalink] ### Show Tags 12 Oct 2018, 08:35 => In order for the sum to be even, both primes selected must be odd. As 2 is the only even prime number, the number of selections with an even sum is equal to the number of ways to select 2 numbers from these 7 odd prime numbers, or 7C2. The total number of selections of 2 prime numbers from the first 8 prime numbers is 8C2. Therefore, the probability that the sum of the two numbers selected is even is 7C2 / 8C2 = $${\frac{(7*6)}{(1*2)}}/{\frac{(8*7)}{(1*2)}} = \frac{6}{8} = \frac{3}{4}.$$ Therefore, the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: If 2 numbers are selected from the first 8 prime numbers, what is the  [#permalink]

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13 Oct 2018, 01:48
first 8 prime numbers are$$=2,3,5,7,11,13,17,19$$
sum of the two prime number will be even when 2 is not included in the pair.
So probability of 2 numbers with sum even$$=\frac{7C2}{8C2}=\frac{21}{28}=\frac{3}{4}$$
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Re: If 2 numbers are selected from the first 8 prime numbers, what is the  [#permalink]

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13 Oct 2018, 18:16
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If 2 numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 2 numbers selected is an even number?

$$A. \frac{1}{2}$$
$$B. \frac{1}{3}$$
$$C. \frac{2}{3}$$
$$D. \frac{1}{4}$$
$$E. \frac{3}{4}$$

The first 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19. We see that all of them are odd numbers except 2, and, in order for the sum of the 2 numbers selected to be even, the two numbers must be odd. Therefore, the probability is:

7/8 x 6/7 = 6/8 = 3/4

Answer: E
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Re: If 2 numbers are selected from the first 8 prime numbers, what is the   [#permalink] 13 Oct 2018, 18:16
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