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Ms. Barton has four children. You are told correctly that

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Ms. Barton has four children. You are told correctly that [#permalink] New post 02 Jun 2003, 11:55
Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys?

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11
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 [#permalink] New post 03 Jun 2003, 02:54
The answer is A.
Since we know for sure that there are at least two girls out of four children, the question can be rephrased as "what is the probability of having two sons out of two children?" Assuming that both son and daughter are equally likely, we have the answer as 1/4
QED
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 [#permalink] New post 03 Jun 2003, 07:33
I dont know the official answer but I think it's A
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 [#permalink] New post 03 Jun 2003, 21:11
In my view, I opt for A
The reason supporting my belief is that

there are possibly 16 ways (2x2x2x2). Assuming that both son and daughter are equally likely, thus we igonore the order. so we divide with 2


16/2 = 8

The possibility that parents have at least two boys =
GGBB
BBGG
GBBG
BGGB

However, because we don't care about the order, thus 4/2 = 2

Eventually, 2 / 8 = 1/4


right ?
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I think its A too!!!! [#permalink] New post 06 Jun 2003, 11:39
I think its A too........I hope its A!!!!!!!!!!!fingers crossed...........
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 [#permalink] New post 27 Jan 2004, 19:41
I think this Q has been discussed already.

Total COmbinations = C of 2 Girls + C of 3 Girls + C of 4 Girls
= 4C2 + 4C3 + 4C4 = 11

Combinations for 2 boys = 4C2 = 6

so P = 6/11
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 [#permalink] New post 28 Jan 2004, 07:44
anandnk,

good shot.

I do really appreciate this as valuable exp, as intially I have gone with A.
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 [#permalink] New post 28 Jan 2004, 08:21
Hi anandnk,

dont u think 4C3 and 4C4 shouldbeadded as that would mean 3 girls and 4 girls, which is not an option since outa 4 children 2 should be boys ?
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 [#permalink] New post 28 Jan 2004, 08:36
Probability = desired_events/Total_events

total_events = 2G + 3G + 4G = 2B + 1B + 0B

This is a tricky question boss. The initial statement Ms Barton has two girls misleads you. If she has 4 children and two are girls then definitely other two are boys ( assuming no in between ). This is indirect way of asking what is the probability that out of 4 children 2 are boys.

I hope it is clear.

Last edited by anandnk on 30 Jan 2004, 07:43, edited 2 times in total.
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 [#permalink] New post 28 Jan 2004, 08:41
Got it !

Thanks anandnk.
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 [#permalink] New post 29 Jan 2004, 15:26
:oops:

the answer is D because its soooo very simple....the probability that out of the other two children (we already know that two are girls) both are boys is 1/2 as in case of any one of the two being a girl the probability does not hold
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 [#permalink] New post 01 Feb 2004, 00:52
Firstly I thought that it is A. But after anandk's explanations I have some doubt. But still don't understand his explanation fully.
We were told that 2 out of 4 children are girls. Which left us with the question " What is the probability of 2 of the rest children are boys.
Probability that 3d child is boy 1/2 so probability of 4th child is boy also 1/2. By using formula we got 1/2*1/2=1/4.
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 [#permalink] New post 01 Feb 2004, 01:03
anandnk wrote:
I think this Q has been discussed already.

Total COmbinations = C of 2 Girls + C of 3 Girls + C of 4 Girls
= 4C2 + 4C3 + 4C4 = 11

Combinations for 2 boys = 4C2 = 6

so P = 6/11


Would you mind to list, say, C of 2 girls?
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 [#permalink] New post 01 Feb 2004, 04:04
Ok. Where am I going wrong here.

I got:

4C2 x (1/2)^2 x (1/2)^2

= 6 x 1/16 = 3/8
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 [#permalink] New post 01 Feb 2004, 06:20
jumbie wrote:
Ok. Where am I going wrong here.

I got:

4C2 x (1/2)^2 x (1/2)^2

= 6 x 1/16 = 3/8


What you have find out is simple probabilty for *EXCATLY* 2 boys and
2girls out of 4 children.

What has been asked is to find conditional probability.
Condition: 4 children and atleast 2 are girls. In other words, there can be
2 boys and 2 girls, 3 girls and 1 boy and all 4 girls.

What has been asked is that
Out of the above mentioned combinations what is the probability of
getting 2 boys and 2girls.

In math teams,

P (2B&2G given that atleast 2 out of 4 are girls) =
Combination(2B & 2G ) / ( (comb(2B+2G) + comb(3G+1B) + comb(4G) )
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 [#permalink] New post 02 Feb 2004, 11:56
dont u thinkthe answer shoud be 1/2

What is the official answer to this question.o

Since we have to find out the P of 2 boys out ouf 4. We have only once possible combinations 2G and 2B. so we have 2 children to chose from and both of them have to sons
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Re: Tricky!! - Probability [#permalink] New post 03 Feb 2004, 06:11
ABNY2002 wrote:
Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys?

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11


This is a question that I submitted a long time ago.

Anand is correct.

This is a question in "conditional probability".

Two ways to solve it:

1) first reduce the "possible" set to match the condition. We know that barton must have at least two girls. Of the 16 possible arrangements of children from youngest to oldest, only 11 qualify. Hence there are 11 possible arrangement of kids. Of those 11 exactly 6 have 2 boys. Hence, the probability is 6/11.

2) The "formula" for conditional probability is "prob without the condition" divided by "the probabllity of the condition". Hence, the odds of getting exactly 2 boys and 2 girls out of the sixteen possible is: 6/16. (BBGG, BGBG, BGGB, GGBB, GBGB, GBBG.), The probability of the condition (at least 2 girls) is 11/16 (16 less 5: BBBB, BBBG, BGBB, BBGB, GBBB).

6/16 divided by 11/16 = 6/11.
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 [#permalink] New post 03 Feb 2004, 12:56
look at similar problem:-)

http://www.wiskit.com/marilyn/boys.html
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 [#permalink] New post 04 Feb 2004, 00:58
Zhung Gazi wrote:
look at similar problem:-)

http://www.wiskit.com/marilyn/boys.html


Marilyn Von Savant is an idiot.
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MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 04 Feb 2004, 00:58
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