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If there are 15 children in a gym class and 2 children are to be

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If there are 15 children in a gym class and 2 children are to be  [#permalink]

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New post 14 Oct 2018, 02:13
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If there are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration, how many of the children are left-handed?
1) The probability of no left-handed children being chosen is \(\frac{3}{7}\)
2) The number of left-handed children is less than the number of right handed children in the class

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If there are 15 children in a gym class and 2 children are to be  [#permalink]

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New post 14 Oct 2018, 03:47
Abhi077 wrote:
If there are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration, how many of the children are left-handed?
1) The probability of no left-handed children being chosen is \(\frac{3}{7}\)
2) The number of left-handed children is less than the number of right handed children in the class


To help understand what we need to do, we'll translate the given data into equations.
This is a Precise approach.

We'll use 'x' to represent the number of left-handed children

(1) so 3/7 = prob of no left-handed = "ways to choose only right handed children" / "ways to choose children" =
(15-x)C2 / (15C2) = (15 - x)(14 - x) / (15*14).
This is complicated to solve precisely, but there is no need! There is at least one answer to this (quadratic) equation, and since changing x = "the number of left-handed children" must also change the probability of choosing a left-handed child, there can be only one applicable answer (the other one will be negative or too large or a fraction or something else irrelevant).
Sufficient.

(2) So x < 15 - x meaning that x < 7.5. With no additional information, x can be any of {1,2,3,4,5,6,7}.
Insufficient.

(A) is our answer.
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Re: If there are 15 children in a gym class and 2 children are to be  [#permalink]

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New post 16 Oct 2018, 05:03
Thanks for the question, just wanted to check though ; if 3/7 is the probability of right handed, the probability for right handed is 4/7.

I had created an equation of 4/7 * 15C2 to get an answer and choose A.

While as solution given below by DavidExamPal is taking 15-x and deriving the answer, while am not saying it' wrong, I am just not sure why 4/7* 15C2 would be less preferable to the approach adopted by David.

Happy to be advised otherwise :)

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Re: If there are 15 children in a gym class and 2 children are to be  [#permalink]

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New post 16 Oct 2018, 17:19
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Abhi077 wrote:
If there are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration, how many of the children are left-handed?

1) The probability of no left-handed children being chosen is \(\frac{3}{7}\)
2) The number of left-handed children is less than the number of right handed children in the class


Target question: How many of the children are left-handed?

Given: There are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration

Statement 1: The probability of no left-handed children being chosen is 3/7
P(no left-handers among those selected) = P(both right-handers among those selected)
So, P(both right-handers among those selected) = 3/7
In other words, P(1st selection is right-handed AND 2nd selection is right-handed) = 3/7
So, P(1st selection is right-handed) x P(2nd selection is right-handed) = 3/7

Let R = # of right-handers in the group

We can write: R/15 x (R - 1)/14 = 3/7
Simplify to get: (R² - R)/210 = 3/7
Cross multiply to get: 7(R² - R) = 3(210)
Divide both sides by 7 to get: R² - R = 3(30)
Or....: R² - R = 90
Then: R² - R - 90 = 0
Factor: (R - 10)(R + 9) = 0
So, EITHER R = 10 OR R = -9
Since R cannot be negative, we know that R = 10 (i.e., there are 10 right-handers)

If 10 children are right-handed, then 5 children are left-handed
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The number of left-handed children is less than the number of right handed children in the class
There are several scenarios that satisfy statement 2. Here are two:
Case a: There are 4 left-handers and 11 right-handers. In this case, the answer to the target question is there are 4 left-handers
Case b: There are 5 left-handers and 10 right-handers. In this case, the answer to the target question is there are 5 left-handers
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Re: If there are 15 children in a gym class and 2 children are to be  [#permalink]

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New post 16 Oct 2018, 19:07
GMATPrepNow wrote:
Abhi077 wrote:
If there are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration, how many of the children are left-handed?

1) The probability of no left-handed children being chosen is \(\frac{3}{7}\)
2) The number of left-handed children is less than the number of right handed children in the class


Target question: How many of the children are left-handed?

Given: There are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration

Statement 1: The probability of no left-handed children being chosen is 3/7
P(no left-handers among those selected) = P(both right-handers among those selected)
So, P(both right-handers among those selected) = 3/7
In other words, P(1st selection is right-handed AND 2nd selection is right-handed) = 3/7
So, P(1st selection is right-handed) x P(2nd selection is right-handed) = 3/7

Let R = # of right-handers in the group

We can write: R/15 x (R - 1)/14 = 3/7
Simplify to get: (R² - R)/210 = 3/7
Cross multiply to get: 7(R² - R) = 3(210)
Divide both sides by 7 to get: R² - R = 3(30)
Or....: R² - R = 90
Then: R² - R - 90 = 0
Factor: (R - 10)(R + 9) = 0
So, EITHER R = 10 OR R = -9
Since R cannot be negative, we know that R = 10 (i.e., there are 10 right-handers)

If 10 children are right-handed, then 5 children are left-handed
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The number of left-handed children is less than the number of right handed children in the class
There are several scenarios that satisfy statement 2. Here are two:
Case a: There are 4 left-handers and 11 right-handers. In this case, the answer to the target question is there are 4 left-handers
Case b: There are 5 left-handers and 10 right-handers. In this case, the answer to the target question is there are 5 left-handers
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent


Hi Brent

Thank You for your response and as always it' detailed and helpful.

Though, advice what I was seeking was an alternate approach , if you can confirm is correct or not.

Given, probability of no left-handed children being chosen is 3/7 in statement 1.

1). We can derive that prob. of both left handy' is 4/7
2). Because we know prob. is 4/7 and total students available, would you agree that this equation is correct - ( 3/7 divided by 15C2 )
3). The only reason am asking is to reassure myself that 3/7 can indeed be used to create such an equation above, knowing it is the ratio of actual students in class and may just be a fraction representation of the individual left/right handed students.

I hope I have been able to share my point of view, do let me know if you need me to clarify more.

Thanks in advance.
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Re: If there are 15 children in a gym class and 2 children are to be  [#permalink]

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New post 17 Oct 2018, 05:16
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proabhinav wrote:
Hi Brent

Thank You for your response and as always it' detailed and helpful.

Though, advice what I was seeking was an alternate approach , if you can confirm is correct or not.

Given, probability of no left-handed children being chosen is 3/7 in statement 1.

1). We can derive that prob. of both left handy' is 4/7
2). Because we know prob. is 4/7 and total students available, would you agree that this equation is correct - ( 3/7 divided by 15C2 )
3). The only reason am asking is to reassure myself that 3/7 can indeed be used to create such an equation above, knowing it is the ratio of actual students in class and may just be a fraction representation of the individual left/right handed students.

I hope I have been able to share my point of view, do let me know if you need me to clarify more.

Thanks in advance.


We can't conclude point #1 ("We can derive that prob. of both left handy' is 4/7")
If we select 2 children, there are 4 possible outcomes:
A) 1st child is lefty and 2nd child is lefty
B) 1st child is lefty and 2nd child is righty
C) 1st child is righty and 2nd child is lefty
D) 1st child is righty and 2nd child is righty

We know that P(event D) = 3/7, which means P(A or B or C) = 4/7
In other words, we can't say that P(A) = 4/7 (as you have suggested)

From here, we must determine the individual probabilities of P(A), P(B) and P(C)

Does that help?

Cheers,
Brent
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Re: If there are 15 children in a gym class and 2 children are to be   [#permalink] 17 Oct 2018, 05:16
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