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805+ (Hard)|   Probability|      
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Abhi077
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The following are the ages of 20 children with first name initials 14 Oct 2018, 10:58
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B
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95% (hard)

Question Stats:

44% (02:50) correct 56% (03:02) wrong based on 80 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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B
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The following are the ages of 20 children with first name initials 14 Oct 2018, 18:41
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Abhi077
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
Show more


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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The following are the ages of 20 children with first name initials 14 Oct 2018, 21:26
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chetan2u
Abhi077
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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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.
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The following are the ages of 20 children with first name initials 14 Oct 2018, 21:38
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chetan2u
Abhi077
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.
Show more

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
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The following are the ages of 20 children with first name initials 12 Oct 2020, 00:01
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2, 3, 4, 8, 9, 3, 8, 5, 8, 6, 4, 3, 8, 9, 8, 4, 9, 2, 9, and 8 [2's(2), 3's(3), 4's(3), 5's(1), 6's(1), 8's(6), 9's(4)]

Sum: 4 + 9 + 12 + 5 + 6 + 48 + 36 = 120
Total: 20

Average: \(\frac{120 }{ 20 = 6. }\)

Group has '2' children. Total ways: \(^{20}{C_2} = 190\)

The child whose age is the average (arithmetic mean) age of the group of 20 children: This means the age of one of the child has to be '6' and others can be anyone.

Total ways: \(1 * ^{19}{C_1} = 1 * 19\)

Percent chances: \(\frac{19}{190}\) * 100 = 10%

Answer B
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