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In a class with 12 children, q of the children are girls.

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ksharma12
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In a class with 12 children, q of the children are girls. [#permalink] New post   20 Apr 2010, 17:40
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In a class with 12 children, q of the children are girls. Two children will be randomly chosen simultaneously. What is the value of q?

(1) The probability that two girls will be chosen together is 1/11.
(2) The probability that one boy will be chosen and one girl will be chosen is 16/33.
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Re: Two children will be randomly chosen simultaneously. DS Prob [#permalink] New post   20 Apr 2010, 18:26
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ksharma12 wrote:
In a class with 12 children, q of the children are girls. Two children will be randomly chosen simultaneously. What is the value of q?

1) The probability that two girls will be chosen together is 1/11.
2) The probability that one boy will be chosen and one girl will be chosen is 16/33.

Multiple Choice Options:
A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.


(1) \(\frac{q}{12}*\frac{q-1}{12-1}=\frac{1}{11}\) --> \(q(q-1)=12\) --> \(q=4\). Sufficient.

(2) \(2*\frac{q}{12}*\frac{12-q}{12-1}=\frac{16}{33}\) --> \(q(12-q)=32\) --> \(q=4\) or \(q=8\). Two answers, not sufficient.

Answer: A.
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Re: Two children will be randomly chosen simultaneously. DS Prob [#permalink] New post   24 Apr 2010, 10:38
Good DS from probablity ... nice explanation
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Re: In a class with 12 children, q of the children are girls. [#permalink] New post   05 Aug 2014, 06:19
Bunuel wrote:
ksharma12 wrote:
In a class with 12 children, q of the children are girls. Two children will be randomly chosen simultaneously. What is the value of q?

1) The probability that two girls will be chosen together is 1/11.
2) The probability that one boy will be chosen and one girl will be chosen is 16/33.

Multiple Choice Options:
A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.


(1) \(\frac{q}{12}*\frac{q-1}{12-1}=\frac{1}{11}\) --> \(q(q-1)=12\) --> \(q=4\). Sufficient.

(2) \(2*\frac{q}{12}*\frac{12-q}{12-1}=\frac{16}{33}\) --> \(q(12-q)=32\) --> \(q=4\) or \(q=8\). Two answers, not sufficient.

Answer: A.


Bunuel,

If option B states, The probability that a boy and a girl will be chosen is 16/33. then is the below equation correct? In this case we can chose 2 combinations BG or GB.

2* (12-q)c1 * qc1/12c2?

Since, option B states ,The probability that one boy will be chosen and one girl will be chosen is 16/33. ....the only possibility is BG

Please correct my understanding
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Re: In a class with 12 children, q of the children are girls. [#permalink] New post   06 Aug 2014, 06:57
Bunuel wrote:
ksharma12 wrote:
In a class with 12 children, q of the children are girls. Two children will be randomly chosen simultaneously. What is the value of q?

1) The probability that two girls will be chosen together is 1/11.
2) The probability that one boy will be chosen and one girl will be chosen is 16/33.

Multiple Choice Options:
A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.


(1) \(\frac{q}{12}*\frac{q-1}{12-1}=\frac{1}{11}\) --> \(q(q-1)=12\) --> \(q=4\). Sufficient.

(2) \(2*\frac{q}{12}*\frac{12-q}{12-1}=\frac{16}{33}\) --> \(q(12-q)=32\) --> \(q=4\) or \(q=8\). Two answers, not sufficient.

Answer: A.



Bunuel,

I do not get why in 2) \(2*\frac{q}{12}*\frac{12-q}{12-1}\) you multiply by 2. If we pick a girl first then a boy, that's what you wrote, but if we pick a boy first isn't it \(\frac{(12-q)}{12} * \frac{q}{(12-1)}\) which ends up being the same anyway?
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Re: In a class with 12 children, q of the children are girls. [#permalink] New post   12 Aug 2014, 04:36
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saikrishna123 wrote:
Bunuel wrote:
ksharma12 wrote:
In a class with 12 children, q of the children are girls. Two children will be randomly chosen simultaneously. What is the value of q?

1) The probability that two girls will be chosen together is 1/11.
2) The probability that one boy will be chosen and one girl will be chosen is 16/33.

Multiple Choice Options:
A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.


(1) \(\frac{q}{12}*\frac{q-1}{12-1}=\frac{1}{11}\) --> \(q(q-1)=12\) --> \(q=4\). Sufficient.

(2) \(2*\frac{q}{12}*\frac{12-q}{12-1}=\frac{16}{33}\) --> \(q(12-q)=32\) --> \(q=4\) or \(q=8\). Two answers, not sufficient.

Answer: A.


Bunuel,

If option B states, The probability that a boy and a girl will be chosen is 16/33. then is the below equation correct? In this case we can chose 2 combinations BG or GB.

2* (12-q)c1 * qc1/12c2?

Since, option B states ,The probability that one boy will be chosen and one girl will be chosen is 16/33. ....the only possibility is BG

Please correct my understanding


You don't need to multiply by 2 when using combinations approach. So, it should be (12-q)C1*qC1/12C2 = 16/33.
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Re: In a class with 12 children, q of the children are girls. [#permalink] New post   18 Feb 2022, 09:54
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Re: In a class with 12 children, q of the children are girls. [#permalink]
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