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Ms. Barton has four children. You are told correctly that she has at
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HisHo
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Re: Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  20 Jun 2018, 07:11
Since there is at least 2 girls, then we have 3 probabilities:
1) 2 boys & 2 girls:
4!/(2!*2!) = 6

2) 1 boy & 3 girls:
4!/3! = 4

3) 4 girls: 1

So the total probabilities are 6+4+1=11
We need the probability of 2B & 2G = 6/11
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Re: Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  20 Jun 2018, 10:09
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


Ms. Barton has at least 2 girls out the 4 children. We need probability of 2 boys & 2 girls.

We have three cases: GGBB, GGGB & GGGG

GGBB can occur in 4!/2!2! = 6 ways

GGGB can occur in 4!/3! = 4 ways

GGGG can occur in only 1 way

Total # of ways = 6 + 4 + 1 = 11

Probability of GGBB = 6/11

Answer E.


Thanks,
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Re: Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  20 Jun 2018, 11:07
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dprchem wrote:
Why are we getting into arrangements. The question is about probability (no matter in which order). Then the answer should be 1/4. why is it wrong?


Ms. Barton currently has two girls. If the probability of having a girl is equal to the probability of having a boy, what is the probability that her next two children are boys?

Here, the first 2 children must be girls.
Below are all the ways for the first 2 children to be girls:
GGBB
GGBG
GGGB
GGGG
Of the 4 possible outcomes, only the 1 case in blue includes 2 boys.
Thus:
P(the next 2 children are boys) = 1/4.

The posted problem is different.
It asks the following:

Of Ms. Barton's 4 children, at least two are girls. If the probability of having a girl is equal to the probability of having a boy, what is the probability that there are 2 boys among Ms. Barton's 4 children?

Here, the first two children do NOT have to be girls.
The only condition is that AT LEAST 2 of the 4 children must be girls.
As a result, a greater number of outcomes are possible.
Below are all the ways for at least 2 girls to be among the 4 children:

4 girls:
GGGG

3 girls:
GGGB
GGBG
GBGG
BGGG

2 girls:
GGBB
GBGB
GBBG
BGGB
BGBG
BBGG
BBGG

Of the 11 ways to have at least 2 girls, only the 6 cases in blue include 2 boys.
Thus:
P(2 boys are among the 4 children) = 6/11.

Show: ::
E

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Re: Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  30 Jun 2018, 23:00
dprchem wrote:
VeritasPrepKarishma wrote:
forevertfc wrote:
What am I missing here?

The questions clearly states that 2 of the 4 kids are girls. It also clearly asks "what is the probability that the other 2 are boys?"

This gives us a P of 1/2 that a child will be a boy and 1/2 that the child will be a girl. Armed with the fact that having two girls will have no significance on the sex of the other two children, we can treat this as mutually exclusive scenario.

So there is P:1/2 that one of the remaining children is a boy, and P:1/2 that the other child is also a boy. Therefore, the probability would become 1/4.

Am I reading this question wrong?


We are concerned about the 4 selections together, not the individual selection. What this means is that if your logic is correct, the probability of having 4 girls would be the same as the probability of having 2 girls and 2 boys. But is that correct? No it is not.

In how many ways can you have 4 girls?
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next girl, next girl, next girl)

In how many ways can you have 2 girls and 2 boys?
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next girl, next boy, next boy)
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next boy, next girl, next boy)
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next boy, next boy, next girl)
and so on...

There is a sequence to obtaining kids and that needs to be factored in. Many ways will lead to 2 girls and 2 boys hence the probability of obtaining 2 girls and 2 boys will be higher than the probability of obtaining 4 girls. The question clearly tells us that we don't know which 2 are girls so we cannot say that this the case where first child and second child are girls and others are boys. Hence you need to follow the method discussed above: http://gmatclub.com/forum/ms-barton-has ... 58#p991399

It is similar to the coin flipping case. Every coin flip is independent of the other but the probability of obtaining 4 heads in 4 flips is less than the probability of obtaining 2 heads and 2 tails.


Why are we getting into arrangements. The question is about probability (no matter in which order). Then the answer should be 1/4. why is it wrong?


It is wrong because in Your solution You are considering that there are 2 girls. But according to question there are at least 2 girls.
So that why we need to make cases and solve the question.
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Re: Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  09 Jul 2018, 14:13
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)


Hi @VeritasPrepKrishna

I am not able to understand why arrangement matter here?

My solution:

there can be either 2 girls, 3 girls or 4 girls i.e. 3 condition. The only possibility for two boys is when family has two girls and two boys i.e. one condition

so probability will be 1/3.

Can you please correct my understanding.
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Rahaman9969
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Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  17 Dec 2018, 00:47
Atleast can be converted to exactly
2 girls or 3 girls or 4 girls

Total ways =4c2 + 4c3 + 4c4=11
Required ways of having boy = 4c2
So the answer is 6/11

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Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  20 Sep 2019, 08:52
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


Given: 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.

Asked: 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.)

Ms Barton has at least 2 girls
Number of girls = {2,3,4}
Number of boys = {2,1,0}
(Boys, Girls) = {(2,2),(1,3),(0,4)}
Case (2,2)
\(^4C_2 = 6\)
Case (1,3)
\(^4C_1 =4\)
Case (0,4)
\(^4C_0 =1\)

The probability that she also has two boys = \(\frac{6}{(6+4+1)} = \frac{6}{11}\)

IMO E
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Re: Ms. Barton has four children. You are told correctly that she has at [#permalink] New post  18 Mar 2020, 08:13
Hi
we are told that the probability of having a boy is just the same of that of a girl, and we have at least 2 girls. at least 2 girls mean either 2G2B or 3G1B or 4G0B with no preferred scenario over the other. Thus we have each of a probability 1/3. What am I missing here?
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