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Ms. Barton has four children. You are told correctly that she has at
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15 Mar 2011, 22:44
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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 has four children. You are told correctly that she has at
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26 Oct 2011, 20:33
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 has four children. You are told correctly that she has at
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16 Mar 2011, 01:47
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 has four children. You are told correctly that she has at
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16 Mar 2011, 01:35
Assume that there are 4 Ms. Barton's children in a row: 1234. 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 GGBBBGGG 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 has four children. You are told correctly that she has at
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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 has four children. You are told correctly that she has at
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17 Mar 2011, 21:01
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 165=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 has four children. You are told correctly that she has at
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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 has four children. You are told correctly that she has at
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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 has four children. You are told correctly that she has at
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26 Oct 2011, 21:58
4 girls, 0 boys  4C4=1 3 girls, 1 boy4C3=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 has four children. You are told correctly that she has at
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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 has four children. You are told correctly that she has at
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16 Nov 2013, 02:52
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 Karishma I have two questions. 1. when calculating 2 boys and 2 girls, given that there are two girls, why do we multiply 4!/2!2! with (1/2)^4? We already know that two are girls, so shouldn't we multiply only by (1/2)^2? 2. When considering the overall, why do we divide (1) by p(at least two girls)? Why don't we divide it by all available options? Thanks



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Re: Ms. Barton has four children. You are told correctly that she has at
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04 Dec 2013, 20:00
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?



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Re: Ms. Barton has four children. You are told correctly that she has at
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04 Dec 2013, 21:10
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: msbartonhasfourchildrenyouaretoldcorrectlythatsh110962.html?sid=9797da5586e03fc5fd356ef7e6808458#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.
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Re: Ms. Barton has four children. You are told correctly that she has at
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04 Jul 2014, 23:40
Favourable outcome = 2G2B = 4!/2!2! = 6 Total outcomes(We have to modify the total outcomes as it is given that there are atleast 2 girls, so we are not considering the case of having 4B, 3B1G) = 2G2B + 3G1B + 4G = (4!/2!2!) + (4!/3!1!) + (4!/4!) = 6+4+1 = 11 Therefore required probability = 6/11



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Re: Ms. Barton has four children. You are told correctly that she has at
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09 Sep 2014, 12:38
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 Karishma, Perhaps you missed my previous question. I don't understand why when calculating the odds of calculating 2 boys and 2 girls, you multiply by (1/2)^4. We are already told that 2 girls are given... So why not multiply just by the odds of the rest being boys, which is (1/2)^2?



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Re: Ms. Barton has four children. You are told correctly that she has at
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09 Sep 2014, 20:52
ronr34 wrote: 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 Karishma, Perhaps you missed my previous question. I don't understand why when calculating the odds of calculating 2 boys and 2 girls, you multiply by (1/2)^4. We are already told that 2 girls are given... So why not multiply just by the odds of the rest being boys, which is (1/2)^2? Probability of A given B = P(A) (out of all possible cases)/P(B) (out of all possible cases) B will be a subset of all possible cases and A will be a subset of B. When you calculate P(A), you assume all possible cases and then divide by P(B) because the only range you have to consider is P(B). Since P(B) is less than 1 usually, Probability of A given B is usually more than P(A).
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Re: Ms. Barton has four children. You are told correctly that she has at
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05 Sep 2016, 00:42
Whats the significance of the following statement in the question "Assume that the probability of having a boy is the same as the probability of having a girl." ?
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JarvisR wrote: Whats the significance of the following statement in the question "Assume that the probability of having a boy is the same as the probability of having a girl." ? This tells you that the probability of having a boy = Probability of having a girl = 0.5 Another question could say that the probability of having a boy = 0.6 which would mean that the probability of having a girl is 0.4. Look at this solution: msbartonhasfourchildrenyouaretoldcorrectlythatsh110962.html#p991399It uses (1/2) which is 0.5. If the two probabilities were different, we would have used 0.6 and 0.4.
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Re: Ms. Barton has four children. You are told correctly that she has at
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31 May 2017, 23:03
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
1. Since you are told there are at least two girls, you proceed from what is certainly known to calculate the total number of cases. So you eliminate the cases of 1 girl and no girl. The total number of cases will be 165=11 2. Favorable cases are ggbb, gbbg, gbgb, bbgg, bggb, bgbg, a total of 6 3. Probability is 6/11.
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Re: Ms. Barton has four children. You are told correctly that she has at
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19 Jun 2018, 00:02
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/msbartonhas ... 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?




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