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# Ms. Barton

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Ms. Barton [#permalink]  15 Mar 2011, 22:44
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26% (03:21) correct 73% (01:16) wrong based on 0 sessions
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? (Assume that the probability of having a boy is the same as the probability of having a girl.)

(A) 1/4 (B) 3/8 (C) 5/11 (D) 1/2 (E) 6/11
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Re: Ms. Barton [#permalink]  16 Mar 2011, 01:35
3
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Assume that there are 4 Ms. Barton's children in a row: 1-2-3-4. Let show all possible combinations of them being girl(G) or boy(B) giving that there at least two girls. We are seeing each child 1,2,3,4 a s a unique(so, we distinguish between them)
BBGG
BGBG
GBBG
BGGB
GBGB
GGBB

BGGG
GBGG
GGBG
GGGB
GGGG

Overall 11 variants. 6 of them satisfy the condition of 2 boys exactly (they are highlighted in bold)
Therefore the probability is 6/11
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Re: Ms. Barton [#permalink]  16 Mar 2011, 01:47
2
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Onell 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? (Assume that the probability of having a boy is the same as the probability of having a girl.)

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

Let me try a general approach without resorting to actually listing all possibilities (which may not be possible if we need to do similar problem for say exactly 3 out of 6 or 4 out of 8 etc.)

number of ways in which there can be two boys and two girls out of 4 children = 4!/((2!)*(2!)) = 6

Number of ways in which there will be atleast 2 girls out of 4 children = ways for 2 boys and 2 girls (6 as calculated above) + ways for 3 girls and 1 boy (=4!/3! = 4)+ways for 1 all girls (=1), so total of 11 ways.

Required probability = 6/11
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Re: Ms. Barton [#permalink]  16 Mar 2011, 04:20
You need to determine the number of favourable arrangements (i.e. arrangements for which # girls = 2) out of all possible arrangements (i.e. arrangements for which # girls >= 2)

# arrangements for 2 girls and 2 boys (favourable arrangement) = 4! / (2!*2!) = 6
# arrangements for 3 girls and 1 boy = 4! / (3!*1!) = 4
# arrangements for 4 girls and 0 boys = 4! / 4! = 1

Thus, # favourable arrangements / # possible arrangements = 6 / (6 + 4+ 1) = 6/11
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Re: Ms. Barton [#permalink]  17 Mar 2011, 01:21
I liked beyondgmatscore's approach a lot. Kudos !

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Re: Ms. Barton [#permalink]  17 Mar 2011, 21:01
1
KUDOS
I like it too, and can be simplified even further.

There are normally 2*2*2*2=16 combinations.

We are told it is not 4B0G, and not 3B1G. It is clear that these are 1 combination, and 4 combinations (the G could be in either of 4 spots) respectively. So 16-5=11 combinations remain.

Since it is symmetric (p(g)=p(b)=0.5), there are also 5 combinations of 4G0B/3G1B, leaving 6/11.

p.s. Dont know if this is better, I never learnt all the equations with the '!' in them and when to use each, and so I always solve these intuitively.
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Re: Ms. Barton [#permalink]  18 Mar 2011, 10:36
2 girls and 2 boys (favourable arrangement) = 4! / (2!*2!) = 6
3 girls and 1 boy = 4! / (3!*1!) = 4
4 girls and 0 boys = 4! / 4! = 1

favourable arrangements / possible arrangements = 6 / (6 + 4+ 1) = 6/11
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Re: Ms. Barton [#permalink]  22 Mar 2011, 09:24
OA IS E. Thanks guys for your help
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Re: Ms. Barton [#permalink]  26 Oct 2011, 13:24
Hi people,

Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.

I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).

Pls disprove me.... I don see where I am wrong, but I know surely I am.

Thanks
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Re: Ms. Barton [#permalink]  26 Oct 2011, 20:33
1
KUDOS
raghupara wrote:
Hi people,

Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.

I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).

Pls disprove me.... I don see where I am wrong, but I know surely I am.

Thanks

The following is the complete explanation for this question:

You need to find Probability of 2 boys and 2 girls given there are at least 2 girls. This is a conditional probability. You have already been given that there are at least 2 girls. You need to find the probability of 2 boys and 2 girls given this condition.
P(A given B) = P(A)/P(B)

P(2 boys, 2 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four children, any 2 could be boys and the other two would be girls. So you have to account for all arrangements: BGBG, BBGG, BGGB etc

P(atleast 2 girls) = P(2 boys, 2 girls) + P(1 boy, 3 girls) + P(4 girls) = 1 - P(4 boys) - P(3 boys, 1 girl)
You can solve P(atleast 2 girls) in any way you like.
P(1 boy, 3 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
P(4 girls) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(atleast 2 girls) = 3/8 + 1/4 + 1/16 = 11/16

P(2 boys, 2 girls given atleast 2 girls) = P(2 boys, 2 girls) / P(atleast 2 girls) = (3/8)/(11/16) = 6/11

In the above given solutions, all the (1/2)s have been ignored because the probability of a boy is 1/2 and of a girl is 1/2 too. Hence in every case you will get (1/2)*(1/2)*(1/2)*(1/2) which can be ignored. We just need to focus on arrangements in each case. This is similar to the famous coin flipping questions (the probability of 3 Heads and a tail etc) (check out Binomial Probability in Veritas Combinatorics book for an explanation)
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Re: Ms. Barton [#permalink]  26 Oct 2011, 21:58
1
KUDOS
4 girls, 0 boys - 4C4=1
3 girls, 1 boy-4C3=4
2 girls, 2 boys -4C2=6
total- 6+4+1=11

so ,the probability of "2 boys" is 6/11
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Re: Ms. Barton [#permalink]  27 Oct 2011, 00:54
VeritasPrepKarishma wrote:
raghupara wrote:
Hi people,

Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.

I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).

Pls disprove me.... I don see where I am wrong, but I know surely I am.

Thanks

The following is the complete explanation for this question:

You need to find Probability of 2 boys and 2 girls given there are at least 2 girls. This is a conditional probability. You have already been given that there are at least 2 girls. You need to find the probability of 2 boys and 2 girls given this condition.
P(A given B) = P(A)/P(B)

P(2 boys, 2 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four children, any 2 could be boys and the other two would be girls. So you have to account for all arrangements: BGBG, BBGG, BGGB etc

P(atleast 2 girls) = P(2 boys, 2 girls) + P(1 boy, 3 girls) + P(4 girls) = 1 - P(4 boys) - P(3 boys, 1 girl)
You can solve P(atleast 2 girls) in any way you like.
P(1 boy, 3 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
P(4 girls) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(atleast 2 girls) = 3/8 + 1/4 + 1/16 = 11/16

P(2 boys, 2 girls given atleast 2 girls) = P(2 boys, 2 girls) / P(atleast 2 girls) = (3/8)/(11/16) = 6/11

In the above given solutions, all the (1/2)s have been ignored because the probability of a boy is 1/2 and of a girl is 1/2 too. Hence in every case you will get (1/2)*(1/2)*(1/2)*(1/2) which can be ignored. We just need to focus on arrangements in each case. This is similar to the famous coin flipping questions (the probability of 3 Heads and a tail etc) (check out Binomial Probability in Veritas Combinatorics book for an explanation)

Thanks VeritasPrepKarisma, for simplifying it so much.
Kudos for the details!
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Re: Ms. Barton [#permalink]  13 Nov 2011, 14:05
I got E. This question is a variation of the boy girl paradox
Re: Ms. Barton   [#permalink] 13 Nov 2011, 14:05
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