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805+ (Hard)|   Probability|      
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ksharma12
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In a class with 12 children, q of the children are girls. 20 Apr 2010, 17:40
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95% (hard)

Question Stats:

43% (02:24) correct 57% (02:05) wrong based on 222 sessions

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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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Bunuel
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In a class with 12 children, q of the children are girls. 20 Apr 2010, 18:26
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ksharma12
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.
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(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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In a class with 12 children, q of the children are girls. 24 Apr 2010, 10:38
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Good DS from probablity ... nice explanation
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In a class with 12 children, q of the children are girls. 05 Aug 2014, 06:19
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Bunuel
ksharma12
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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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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In a class with 12 children, q of the children are girls. 06 Aug 2014, 06:57
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Bunuel
ksharma12
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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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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In a class with 12 children, q of the children are girls. 12 Aug 2014, 04:36
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saikrishna123
Bunuel
ksharma12
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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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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In a class with 12 children, q of the children are girls. 18 Feb 2022, 09:54
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