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10 Oct 2018, 01:01
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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83% (01:55) correct
17%
(02:23)
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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?
\(A. \frac{1}{2}\)
\(B. \frac{1}{3}\)
\(C. \frac{2}{3}\)
\(D. \frac{1}{4}\)
\(E. \frac{3}{4}\)
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?
\(A. \frac{1}{2}\)
\(B. \frac{1}{3}\)
\(C. \frac{2}{3}\)
\(D. \frac{1}{4}\)
\(E. \frac{3}{4}\)
10 Oct 2018, 01:43
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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
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
10 Oct 2018, 05:42
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MathRevolution
\({\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{\\
\,{\rm{first}} = 2 = {\rm{even}} \hfill \cr \\
\,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,\)
\(? = P\left( {2\,\,{\text{different}}\,\,{\text{selected}}\,\,{\text{have}}\,{\text{sum}}\,\,{\text{even}}} \right) = P\left( {{\text{number}}\,\,{\text{2}}\,\,{\text{is}}\,\,{\text{not }}\,{\text{selected}}} \right)\)
\({\text{total}} = C\left( {8,2} \right)\,\,\,{\text{equiprobable}}\)
\({\text{favorable}}\,{\text{ = }}\,{\text{C}}\left( {7,2} \right)\,\,\,\,\,\,\left[ {{\text{number}}\,{\text{2}}\,\,{\text{is}}\,{\text{not}}\,{\text{an}}\,{\text{option}}} \right]\)
\(? = \frac{{C\left( {7,2} \right)}}{{C\left( {8,2} \right)}} = \frac{{7 \cdot 6}}{{8 \cdot 7}} = \frac{3}{4}\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
