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If three prime number are randomly selected from prime numbers less

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If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 15 Jun 2018, 03:14
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If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%

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If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 15 Jun 2018, 07:10
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Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%

Find the probability of "NOT odd," a task that is much easier than that in the prompt, and subtract from 1.

There are 10 prime numbers that are less than 30:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

9 of 10 are odd numbers

The sum of any three numbers NOT including 2 will be odd.*

The sum of any 3 odds will be odd:
(O + O) = E (first two odd #s), and then
(E + O) (plus third odd #) = Odd
Example: (3 + 5) = 8, and then (8 + 7) = 15

Use the probability complement rule

Even = not Odd
Odd = not Even
P(E) + P(not E) = 1
P(Even Sum) + P(Odd Sum, which is Not E) = 1

P(Even) = 1 - P(Odd)

Probability of odd sum

There are 10 numbers to pick from. 9 are odd.
P of First Pick odd: \(\frac{9}{10}\)

Only 9 numbers remain. 8 are odd.
P of Second Pick odd: \(\frac{8}{9}\)

8 numbers remain. 7 are odd
P of Third Pick odd: \(\frac{7}{8}\)

P(Odd Sum): \((\frac{9}{10}*\frac{8}{9}*\frac{7}{8})=\frac{7}{10}=0.70\)

P(Even Sum) = \((1 - 0.70) = 0.30\)
= 30 percent

Answer C

If need be, try adding numbers to find the pattern:
(2+3+5) = 10 ... EVEN sum
(3+5+7) = 15 ... ODD sum
(5+7+11) = 23 ... ODD sum
(7+11+13) = 31 ... ODD sum
Pattern: 3 odds sum to ODD. Leave 2 out of the picture until the end.
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 15 Jun 2018, 07:13
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Only one prime number is even and all other primes are odd. So for the sum to be even : two primes should be odd and the other one should be even prime.

No. of prime number (2,3,5,7,11,13,17,19, 23, 29) = 10 numbers.

Even prime can be selected in 1C1 ways = 1
Odd prime can be selected in 9C2 ways = 36

Total ways in which 3 numbers can be selected = 10C3 = 120

Therefore percentage of sum of 3 primes being even = (1*36)/120 = (3/10)*100 = 30%

OA - C.
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 24 Aug 2018, 08:05
1
To have the sum as an even number one of the numbers should be even.

ODD+ODD+EVEN=EVEN
ODD+EVEN+ODD=EVEN

Logically we can conclude that the probability that the prime number 2 will be selected out of 9 other odd numbers is relatively low.
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 24 Aug 2018, 12:05
Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%


total possibilities=10*9*8=720
total odd possibilities=9*8*7=504
total even possibilities=720-504=216
216/720=.3=30%
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 24 Aug 2018, 12:49
pr3mk5: Your answer is brilliantly short and intuitive - thanks. It contained the following:
"Even prime can be selected in 1C1 ways = 1
Odd prime can be selected in 9C2 ways = 36

Total ways in which 3 numbers can be selected = 10C3 = 120"

Could you please elaborate more on how to get the "total ways" - > didnt get the "10C3" (actually dont know what's the "C" in there)

Thanks
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 29 Aug 2018, 11:14
There are 10 prime numbers that are less than 30:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Only 2 is even , rest are odd

Now if the sum of three prime numbers is even , 2 must be selected in one of the three number otherwise the result will always be ODD as ODD +ODD+ODD = ODD

Now there are three ways -
1) 2 AND any odd number AND any odd number
OR
2) any odd number AND 2 AND any odd number
OR
3) any odd number AND any odd number AND 2

Probability -
(1/10)*(9/9)*(8/8)
+
(9/10)*(1/9)*(8/8)
+
(9/10)*(8/9)*(1/8)
= 3/10=0.3

Ans C
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If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 30 Aug 2018, 09:17
Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%



29, 23, 19, 17, 13, 11, 7, 5, 2

Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(9-3)} = 84\)

Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(9-2)} = 36\)

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

\(7C1 = \frac{7!}{(7-1)}= 7\)

\(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\)

So probability is \(\frac{43}{84}\) = ca. \(0.5\) :?


hello there :) pushpitkc - what, where and why ? :-) it looks like a logicaly "correct" solution but i got incorrect answer :) what did i do wrong :?
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If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 30 Aug 2018, 10:39
1
dave13 wrote:
Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%



29, 23, 19, 17, 13, 11, 7, 5, 2

Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(9-3)} = 84\)

Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(9-2)} = 36\)

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

\(7C1 = \frac{7!}{(7-1)}= 7\)

\(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\)

So probability is \(\frac{43}{84}\) = ca. \(0.5\) :?


hello there :) pushpitkc - what, where and why ? :-) it looks like a logicaly "correct" solution but i got incorrect answer :) what did i do wrong :?


Hey dave13

You did everything right except for this - there are 10 prime numbers less than 30.

The total possibilities will change accordingly. This is what the correct solution would look like :)

Listing all primes less than 30 - 29,23,19,17,13,11,7,5,3,2

Total ways of selecting 3 numbers from 10:\(C_3^{10} = \frac{10*9*8}{3*2*1} = 120\)

There is only one possibility to choose 3 prime numbers and have an even sum,
when we have 2 odd and 1 even prime number as the 3 primes need to be different

Total possibility of choosing 2 odd primes from 9: \(C_3^{9} = \frac{9*8}{2*1} = 36\)

There is only one possibility of choosing an even prime number. Therefore, Probability is \(\frac{36}{120} = 30\)%(Option C)

Hope this clears your confusion!
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 01 Sep 2018, 06:02
judsonvpires wrote:
pr3mk5: Your answer is brilliantly short and intuitive - thanks. It contained the following:
"Even prime can be selected in 1C1 ways = 1
Odd prime can be selected in 9C2 ways = 36

Total ways in which 3 numbers can be selected = 10C3 = 120"

Could you please elaborate more on how to get the "total ways" - > didnt get the "10C3" (actually dont know what's the "C" in there)

Thanks


judsonvpires
C stands for combinations

\(nCk = \frac{n!}{(n-k)!*k!}\)

\(10C3 = \frac{10!}{(10-3)!*3!} =\frac{10!}{7!*3!}\)
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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 02 Sep 2018, 05:43
dave13 wrote:
Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%



29, 23, 19, 17, 13, 11, 7, 5, 2

Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(9-3)} = 84\)

Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(9-2)} = 36\)

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

\(7C1 = \frac{7!}{(7-1)}= 7\)

\(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\)

So probability is \(\frac{43}{84}\) = ca. \(0.5\) :?


hello there :) pushpitkc - what, where and why ? :-) it looks like a logicaly "correct" solution but i got incorrect answer :) what did i do wrong :?


The simple logic here is to understand that in reality there are two cases here. As we have only one even number in the list of prime numbers (only 2), there can be only two cases to choose three prime numbers:
1st case: odd+odd+odd
2nd case: odd+odd+even
So, to make our calculations easier, we can subtract the 1st case from the total or 1 to get the 2nd case.

In order to calculate the possibilities of the 1st case, we should do a simple probability calculation. As there are ten prime numbers in that range: 2,3,5,7,11,13,17,19,23,29.
9/10*8/9*7/8=504/720
And finally subtract 504/720 from the total, 1 to get 0.3, which in percent form is 30%.
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If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 02 Sep 2018, 08:36
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Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%


There are 10 prime numbers below 30 and only one such number is even. For the sum of three selected numbers to be even, we have to select the even prime number (i.e. 2). If we don't select it, the sum will be odd. And if we select it, doesn't matter which other numbers we choose, the sum will be even.

So, essentially, the probability asked in the question is the same as the probability that we'll select 2 as one of the three numbers.

So, we have to select 2 as one of the numbers and the other two numbers can be any of the other 9 prime numbers. So, total ways to select such a set of three numbers out of a set of 10 prime numbers is 1C1 * 9C2.

Total number of ways to select 3 numbers out of 10 numbers is 10C3.

The required probability is \(\frac{9C2}{10C3}\) i.e. 0.3.

I hope it helps.

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Re: If three prime number are randomly selected from prime numbers less  [#permalink]

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New post 03 Sep 2018, 17:59
Princ wrote:
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?

A) 10%
B) 27%
C) 30%
D) 36.5%
E) 42%


The prime numbers less than 30 are:

2, 3, 5, 7, 11, 13, 17, 19, 23 and 29

The number of ways of picking any 3 of them is 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2) = 120.

Our goal is to add 3 primes to obtain an even sum. Note that if all 3 prime numbers are odd, we will never obtain an even sum, so we know that one of the 3 primes chosen must be the number 2. So, from the remaining 9 primes, we need to pick any two of them. The number of ways of picking 2 primes from 9 is 9C2 = 9!/(2! x 7!) = (9 x 8)/2 = 36. Therefore, the probability is 36/120 = 3/10 = 30%.

Answer: C
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Re: If three prime number are randomly selected from prime numbers less &nbs [#permalink] 03 Sep 2018, 17:59
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