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A couple decides to have 4 children. If they succeed in havi
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13 Mar 2014, 02:29
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The Official Guide For GMAT® Quantitative Review, 2ND EditionA couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys? (A) 3/8 (B) 1/4 (C) 3/16 (D) 1/8 (E) 1/16 Problem Solving Question: 160 Category: Arithmetic Probability Page: 83 Difficulty: 700 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: A couple decides to have 4 children. If they succeed in havi
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Re: A couple decides to have 4 children. If they succeed in havi
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25 Apr 2014, 01:06
no of ways of getting P(GGBB) is 4!/2!*2!; Total no of ways is 2^n =2^4 =16; 6/16 = 3/8; We can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads . P(all out comes) = 1/2 *1/2 *1/2 *1/2 =1/16; P(favorable outcomes) = 4!/(2! * 2!) = 6/16 =3/8; (OR) Second Approach GBGB GGBB BBGG BGBG GBBG BGGB 6 possible ways . total no of ways is Baby can be a boy or a girl. For each baby the probability is 1/2 ;for 4 babies it's 1/16; 6/16 = 3/8; Press Kudos if this helps




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Re: A couple decides to have 4 children. If they succeed in havi
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15 Mar 2014, 10:59



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Re: A couple decides to have 4 children. If they succeed in having 4 child
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10 Aug 2014, 02:40
By fundamental counting principle, Total No of out comes: 2^4 = 16 Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6 Probability is 6/16 = 3/8
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Re: A couple decides to have 4 children. If they succeed in havi
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Updated on: 15 Mar 2015, 00:07
So here we have four positions and there are two options to fill each position so total number of cases=2x2x2x2=16 now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl favourable cases=4C2=6 probability=6/16=3/8 (A)
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Re: A couple decides to have 4 children. If they succeed in havi
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14 Mar 2015, 08:14
1/2 is the chance of boy or girl
1/2 * 1/2 * 1/2 * 1/2 = 1/16 (4 children, boy or girl)
Possible ways of 2 boys, 2 girls:
GGBB BBGG GBGB BGBG BGGB GBBG
= 6 ways
we need "OR" > 1/16 OR 1/16 .... 1/16 + 1/16 + 1/16 ....... +1/16 = 6/16 = 3/8
Answer A



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Re: A couple decides to have 4 children. If they succeed in having 4 child
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16 Jul 2015, 20:06
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG  since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery. We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls. So shouldn't the probability be 1/5? too simplistic  I know. but where am I wrong?



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Re: A couple decides to have 4 children. If they succeed in havi
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03 Nov 2015, 21:17
Binomial Probability The probability of achieving exactly k successes in n trials is shown below. Formula: P(Probability of K successes in n trials) = nCk p^k q^nk
n = number of trials k = number of successes n – k = number of failures p = probability of success in one trial q = 1 – p = probability of failure in one trial
According to our question
n(4 children) = 4 k( we want exactly 2 girls) = 2 n – k = 2 p (probability of getting a girl in one trial) = 1/2 q = 1 – p = 1/2
4C2 * 1/2^2 * 1/2^2 =3/8
Binomial can also be used for problem like coins.
Binomial is used when following conditions are satisfied.
Fixed number of trials Independent trials Two different classifications Probability of success stays the same for all trials



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Re: A couple decides to have 4 children. If they succeed in havi
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04 Nov 2015, 01:08
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionA couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys? (A) 3/8 (B) 1/4 (C) 3/16 (D) 1/8 (E) 1/16 Let us handle a single case. First boy, second boy, third girl, fourth girl. The probability of this case is 1/2*1/2*1/2*1/2 = 1/16 Two boys or two girls can be born among 4 children in 4C2 ways= 6 ways So the required probability is: total number of ways * probability of one such way= 6* (1/16)=3/8
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Re: A couple decides to have 4 children. If they succeed in having 4 child
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04 Feb 2016, 16:50
donkadsw wrote: Why are we ordering here? I mean, I would think BBGG to be the same as BGBG  since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery. We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls. So shouldn't the probability be 1/5? too simplistic  I know. but where am I wrong? Bump seeking an answer to the question above: I got 1/5 as well. BBBB BBBG BBGG GGGB GGGG 1/5 outcomes The question doesn't say anything about the order. How do we know when order matters?



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Re: A couple decides to have 4 children. If they succeed in having 4 child
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04 Feb 2016, 22:35
Dondarrion wrote: donkadsw wrote: Why are we ordering here? I mean, I would think BBGG to be the same as BGBG  since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery. We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls. So shouldn't the probability be 1/5? too simplistic  I know. but where am I wrong? Bump seeking an answer to the question above: I got 1/5 as well. BBBB BBBG BBGG GGGB GGGG 1/5 outcomes The question doesn't say anything about the order. How do we know when order matters? In this case we are saying that the probability of BGGG is the same as the probability of BBGG. But that is not so. You can have 1 boy and 3 girls in 4 ways: BGGG, GBGG, GGBG, GGGB But you can have 2 boys and 2 girls in 6 ways: BBGG, BGGB, GGBB, BGBG, GBGB, GBBG So the probability depends on the number of ways in which you can get 2 boys and 2 girls. Think of it this way: if you throw two dice, is the probability of getting a sum of 2 same as the probability of getting a sum of 8? No. For sum fo 2, you must get 1 + 1 only. For sum of 8, you could get 4 + 4 or 3 + 5 or 2 + 6 etc. So probability of getting sum of 8 would be higher. In the same way, the order matters in this question.
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Re: A couple decides to have 4 children. If they succeed in having 4 child
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06 Aug 2016, 15:17
Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help?



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Re: A couple decides to have 4 children. If they succeed in having 4 child
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08 Aug 2016, 03:22
ManonZ wrote: Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help? Do you mean in the total number of cases? If yes, then it is actually 2^4. First child can happen in 2 ways (boy or girl). Second, third and fourth kids can also happen in 2 ways. Total number of ways = 2*2*2*2 = 16
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Re: A couple decides to have 4 children. If they succeed in havi
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23 Feb 2018, 13:22
Hi All, There are a couple of ways to answer this question: 1) You could make a table of all the options (there are only 16 possible outcomes) and count up all the ways to get 2 boys and 2 girls or 2) You can do the math Here's the math approach: Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question... 4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls Final Answer = GMAT assassins aren't born, they're made, Rich
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Re: A couple decides to have 4 children. If they succeed in havi
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18 Mar 2018, 11:27
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide. I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong. Could you explain how you determined it to be 600 level difficulty? Thanks, Bunuel! You're the best



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Re: A couple decides to have 4 children. If they succeed in havi
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18 Mar 2018, 12:07



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Re: A couple decides to have 4 children. If they succeed in havi
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07 May 2018, 10:28
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionA couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys? (A) 3/8 (B) 1/4 (C) 3/16 (D) 1/8 (E) 1/16 We need to determine the probability of P(BBGG). P(BBGG) = (1/2)^4 = 1/16 The number of ways to arrange BBGG is 4C2 = 4!/(2! x 2!) = 3 x 2 = 6. Thus, the total probability is 6/16 = 3/8. Answer: A
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