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Math Revolution GMAT Instructor V
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If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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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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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) = $$\frac{21}{56}$$= $$\frac{3}{8}$$
E
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If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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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

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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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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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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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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[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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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.

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Math Revolution GMAT Instructor V
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Re: If 3 different numbers are selected from the first 8 prime numbers, wh  [#permalink]

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=>

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

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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 Re: If 3 different numbers are selected from the first 8 prime numbers, wh   [#permalink] 04 Oct 2018, 01:41
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