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If 3 different numbers are selected from the first 8 prime numbers, wh

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If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 00:37
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[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)

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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 01:02
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)



we cannot have all 3 odd numbers
because o+o+o =o
therefore, one even prime number , which is 2 is always required .
no of ways we can select 2 = 1
no of ways we can select the remaining = \(7c2\) = 21
no of ways we can select 3 = \(8c3\) = 56
p(sum is even) = \(\frac{21}{56}\)= \(\frac{3}{8}\)
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If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 07:00
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)


The first 8 prime numbers are: {2, 3, 5, 7, 11, 13, 17, 19}
Notice that only one prime number is EVEN, and the remaining seven numbers are ODD.
Also notice that the sum of 3 numbers will be EVEN only if one of the three selected numbers is 2
So, the question is really asking "What is the probability that 2 is among the three selected values?"

There are many ways to answer this question.
A super-quick approach is to use the complement
That is, P(Event A happening) = 1 - P(Event A not happening)
So, we can write: P(2 is among the three selected values) = 1 - P(2 is NOT among the three selected values)

P(2 is NOT among the three selected values)
P(2 is NOT among the three selected values) = P(all three selected values are ODD)
= P(1st number is odd AND 2nd number is odd AND 3rd number is odd)
= P(1st number is odd) x P(2nd number is odd) x P(3rd number is odd)
= 7/8 x 6/7 x 5/6
= 5/8

So, P(2 is among the three selected values) = 1 - 5/8
= 3/8

Answer: E

Cheers,
Brent
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 10:01
Since the sum is even , it must contain 2 as one of the numbers. Remaining numbers can be selected in 7c2 ways
Probability = 7c2/8c3 = 3/8
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 10:01
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)


First 8 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19.
The sum will be even if 2 is among the three numbers selected, with the result that the sum = EVEN PRIME + ODD PRIME + ODD PRIME = EVEN.
Since 3 NUMBERS are selected from 8 OPTIONS, the probability that 2 will be among the three numbers selected = \(\frac{3}{8}\).


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If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 11:09
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)

\({\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{
\,{\rm{first}} = 2 = {\rm{even}} \hfill \cr
\,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,\)

\(? = {\mathop{\rm P}\nolimits} \left( {{\rm{3}}\,\,\underline {{\rm{different}}} \,\,{\rm{selected}}\,\,{\rm{have}}\,\,{\rm{sum}}\,\,{\rm{even}}} \right) = {\mathop{\rm P}\nolimits} \left( {{\rm{number}}\,{\rm{2}}\,\,{\rm{is}}\,\,{\rm{selected}},\,\,{\rm{other}}\,\,2\,\,{\rm{free}}} \right)\)


\({\rm{total}} = C\left( {8,3} \right)\,\,\,{\rm{equiprobable}}\)

\({\rm{favorable}}\,\,{\rm{ = }}\,\,{\rm{C}}\left( {7,2} \right)\,\,\,\,\left[ {2\,\,{\rm{had}}\,\,{\rm{already}}\,\,{\rm{stolen}}\,\,{\rm{a}}\,{\rm{place,}}\,\,{\rm{have}}\,\,{\rm{7}}\,\,{\rm{options}}\,\,{\rm{for}}\,\,{\rm{the}}\,{\rm{ other}}\,\,{\rm{two }}\,{\rm{numbers}}\,} \right]\)

\(? = {{{{7!} \over {2!\,\,5!}}} \over {{{8!} \over {3!\,\,5!}}}} = {{7!\,\,3!} \over {2!\,\,8!}} = {3 \over 8}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 02 Oct 2018, 12:58
[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)

total possibilities: 8*7*6=336
odd sum possibilities: 7*6*5=210
even sum possibilities: 336-210=126
126/336=3/8
E
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 03 Oct 2018, 17:00
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even?

\(A. \frac{1}{4}\)
\(B. \frac{1}{3}\)
\(C. \frac{1}{2}\)
\(D. \frac{2}{3}\)
\(E. \frac{3}{8}\)


We recall that odd + odd + odd = odd and that even + odd + odd = even.

The first 8 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19. There are 8C3 = (8 x 7 x 6)/(3 x 2) = 56 ways to choose 3 of them. However, in order for the sum of the 3 numbers to be even, one of them must be 2 (the other two numbers can be any two of the remaining 7 odd primes). The number of ways of choosing 2 odd primes from 7 odd primes is 7C2 = (7 x 6)/2 = 21. Therefore, the probability that the sum of the 3 numbers selected is even is 21/56 = 3/8.

Answer: E
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 03 Oct 2018, 23:57
=>

Suppose \(p, q\) and \(r\) are prime numbers.
In order for \(p + q + r\) to be even, one of them must equal \(2\), since \(2\) is the only even prime number.
Once \(2\) has been selected, there are \(7\) prime numbers remaining from which to select \(2\) numbers. Thus, the number of selections in which the sum of the \(3\) numbers is even is 7C2 = 21.
The total number of selections is 8C3 = 56.
Thus, the probability that the sum of the three numbers is even is \(\frac{21}{56} = \frac{3}{8}.\)

Therefore, the answer is E.
Answer: E
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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New post 04 Oct 2018, 00:41
I went with complement method.
Probability of all odd numbers being selected= 7/8*6/7*5/6=5/8--- 5/8th probability that sum of selected numbers is odd
Probability of selecting the only even prime number= 1-5/8= 3/8 the probability that sum of selected number is even
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh &nbs [#permalink] 04 Oct 2018, 00:41
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