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SC Moderator V
Joined: 25 Sep 2018
Posts: 872
Location: United States (CA)
Concentration: Finance, Strategy
GPA: 3.97
WE: Investment Banking (Investment Banking)
The following are the ages of 20 children with first name initials  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 43% (02:38) correct 57% (02:44) wrong based on 32 sessions

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The following are the ages of 20 children with first name initials from A to T, respectively, in a privately owned school: 2, 3, 4, 8, 9, 3, 8, 5, 8, 6, 4, 3, 8, 9, 8, 4, 9, 2, 9, and 8. If a group consisting of 2 children is to be formed, what is the probability that a group will be selected that contains a child whose age is the average (arithmetic mean) age of the group of 20 children?

A) 12.5 percent
B) 10 percent
C) 7.5 percent
D) 5 percent
E) 2.5 percent

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Math Expert V
Joined: 02 Aug 2009
Posts: 8753
Re: The following are the ages of 20 children with first name initials  [#permalink]

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Abhi077 wrote:
The following are the ages of 20 children with first name initials from A to T, respectively, in a privately owned school: 2, 3, 4, 8, 9, 3, 8, 5, 8, 6, 4, 3, 8, 9, 8, 4, 9, 2, 9, and 8. If a group consisting of 2 children is to be formed, what is the probability that a group will be selected that contains a child whose age is the average (arithmetic mean) age of the group of 20 children?

A) 12.5 percent
B) 10 percent
C) 7.5 percent
D) 5 percent
E) 2.5 percent

So the first thing is to find the average age and that can lead to missing out on some number as there are 20 integers..
So let's find number of each digit..
2 - 2 times
3- 3 times
4- 3 times
5- 1 times
6- 1 times
8- 6 times
9- 4 times
Total numbers - 2+3+3+1+1+5+4=20, so ok
Average = $$\frac{2*2+3*3+4*3+5+6+6*8+9*4}{20}=\frac{4+9+12+5+6+48+36}{20}=\frac{120}{20}$$=6
So only one has the average age of 6..
Ways this child will be a part of two choose = 1*19
Ways to choose 2 out of 20 =20C2=190
So probability = 19/190=1/10 or 10%

B
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Re: The following are the ages of 20 children with first name initials  [#permalink]

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chetan2u wrote:
Abhi077 wrote:
The following are the ages of 20 children with first name initials from A to T, respectively, in a privately owned school: 2, 3, 4, 8, 9, 3, 8, 5, 8, 6, 4, 3, 8, 9, 8, 4, 9, 2, 9, and 8. If a group consisting of 2 children is to be formed, what is the probability that a group will be selected that contains a child whose age is the average (arithmetic mean) age of the group of 20 children?

A) 12.5 percent
B) 10 percent
C) 7.5 percent
D) 5 percent
E) 2.5 percent

So the first thing is to find the average age and that can lead to missing out on some number as there are 20 integers..
So let's find number of each digit..
2 - 2 times
3- 3 times
4- 3 times
5- 1 times
6- 1 times
8- 6 times
9- 4 times
Total numbers - 2+3+3+1+1+5+4=20, so ok
Average = $$\frac{2*2+3*3+4*3+5+6+6*8+9*4}{20}=\frac{4+9+12+5+6+48+36}{20}=\frac{120}{20}$$=6
So only one has the average age of 6..
Ways this child will be a part of two choose = 1*19
Ways to choose 2 out of 20 =20C2=190
So probability = 19/190=1/10 or 10%

B

Can you please elaborate how you got this:Ways this child will be a part of two choose = 1*19
I did the remaining solution just like stated,but failed at this step.
Math Expert V
Joined: 02 Aug 2009
Posts: 8753
Re: The following are the ages of 20 children with first name initials  [#permalink]

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chetan2u wrote:
Abhi077 wrote:
The following are the ages of 20 children with first name initials from A to T, respectively, in a privately owned school: 2, 3, 4, 8, 9, 3, 8, 5, 8, 6, 4, 3, 8, 9, 8, 4, 9, 2, 9, and 8. If a group consisting of 2 children is to be formed, what is the probability that a group will be selected that contains a child whose age is the average (arithmetic mean) age of the group of 20 children?

A) 12.5 percent
B) 10 percent
C) 7.5 percent
D) 5 percent
E) 2.5 percent

So the first thing is to find the average age and that can lead to missing out on some number as there are 20 integers..
So let's find number of each digit..
2 - 2 times
3- 3 times
4- 3 times
5- 1 times
6- 1 times
8- 6 times
9- 4 times
Total numbers - 2+3+3+1+1+5+4=20, so ok
Average = $$\frac{2*2+3*3+4*3+5+6+6*8+9*4}{20}=\frac{4+9+12+5+6+48+36}{20}=\frac{120}{20}$$=6
So only one has the average age of 6..
Ways this child will be a part of two choose = 1*19
Ways to choose 2 out of 20 =20C2=190
So probability = 19/190=1/10 or 10%

B

Can you please elaborate how you got this:Ways this child will be a part of two choose = 1*19
I did the remaining solution just like stated,but failed at this step.

So we have two choose 2 out of 20 but one is already decided..
So first can be only one, whose weight is 6 and the second can be chosen from remaining (20-1)=19
So 1*19
_________________ Re: The following are the ages of 20 children with first name initials   [#permalink] 14 Oct 2018, 20:38

# The following are the ages of 20 children with first name initials  