December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL. December 14, 2018 December 14, 2018 09:00 AM PST 10:00 AM PST 10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 25 Dec 2009
Posts: 78

The probability that a family with 6 children has exactly
[#permalink]
Show Tags
Updated on: 10 Feb 2012, 11:42
Question Stats:
59% (01:25) correct 41% (01:33) wrong based on 833 sessions
HideShow timer Statistics
The probability that a family with 6 children has exactly four boys is: A. 1/3 B. 1/64 C. 15/64 D. 3/8 E. none of the above
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by GMATMadeeasy on 08 Jan 2010, 16:56.
Last edited by Bunuel on 10 Feb 2012, 11:42, edited 1 time in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: probability Qs.. attention
[#permalink]
Show Tags
08 Jan 2010, 22:35
The probability that a family with 6 children has exactly four boys is: A. 1/3 B. 1/64 C. 15/64 D. 3/8 E. none of the above NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)time (6 in our case) sequence is: \(P = C^k_n*p^k*(1p)^{nk}\) \(P = C^k_n*p^k*(1p)^{nk}= C^4_6*(\frac{1}{2})^4*(1\frac{1}{2})^{64}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\) Consider this: We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\). Answer: C.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Manager
Joined: 27 Apr 2008
Posts: 173

Re: probability Qs.. attention
[#permalink]
Show Tags
08 Jan 2010, 19:15
Is it E? Here is my reasoning:
Now let's find the # of combinations where there's a BGB combination. Let the sequence of BG=A: BBAA BAAB BABA ABBA ABAB AABB
So there are 6 combinations here. But since BG can also be GB, we have to multiply by 2. We have to multiply by another 2 because there are 2 As. Therefore, there are 6x2x2=24 combinations.
We also have to include the instances where GG occurs at the end or the front, that is: GGBBBB BBBBGG (all the other instances such as BBGGBB have been taken care of in the calculation at the top)
So we have 24+2=26 possibilities of 4 boys and 2 girls.
The total number of possibilities is 2^6=64. Therefore, the probability is 26/64, which is 13/32. The answer is E, none of the above.



Manager
Joined: 25 Dec 2009
Posts: 78

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 03:11
mrblack wrote: Is it E? Here is my reasoning:
Now let's find the # of combinations where there's a BGB combination. Let the sequence of BG=A: BBAA BAAB BABA ABBA ABAB AABB
So there are 6 combinations here. But since BG can also be GB, we have to multiply by 2. We have to multiply by another 2 because there are 2 As. Therefore, there are 6x2x2=24 combinations.
We also have to include the instances where GG occurs at the end or the front, that is: GGBBBB BBBBGG (all the other instances such as BBGGBB have been taken care of in the calculation at the top)
So we have 24+2=26 possibilities of 4 boys and 2 girls.
The total number of possibilities is 2^6=64. Therefore, the probability is 26/64, which is 13/32. The answer is E, none of the above. Your solution can be simplified this way as well: Total number of events : 2^6 as each position can be filled in exactly two ways Events in favor : (4 Boys and Two Girls; It is like 6 objects in which 4 are alike and 2 others are alike) so 6!/2!4! If it were a probability question, I will follow Bunuel's aproach but the solution above is also important if it is counting problem. Question: If we have n objects out f which p are alike, q are alike, how many possible number of comnitations are possible ? If it were a permutation problem, answer is n!/p!*q! as in the question above



Intern
Joined: 20 Dec 2009
Posts: 12

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 04:18
I think this way:
For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes. Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways. Now here I don't think that ordering does matter in this case.
So the probability comes out to be 15/64.
P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems.



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 04:56
ashueureka wrote: I think this way:
For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes. Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways. Now here I don't think that ordering does matter in this case.
So the probability comes out to be 15/64.
P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems. This way of solving is also correct. You've already considered all possible arrangements for BBBBGG with 6C4, which is 6!/4!2!. If you look at my solution and at yours you'll see that both wrote the same formulas but with different approach.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 06:07



Manager
Status: Its Wow or Never
Joined: 11 Dec 2009
Posts: 165
Location: India
Concentration: Technology, Strategy
Schools: Kelley '14, Tepper '14, KenanFlagler '14, ISB, Goizueta '14, Tippie, Georgia Tech, Katz, Schulich, Mays, Smith '15
GMAT 1: 670 Q47 V35 GMAT 2: 710 Q48 V40
WE: Information Technology (Computer Software)

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 07:06
hi bunuel could u plz provide a few application qs. to this formula? TIA Bunuel wrote: NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1p)^{nk}\)
\(P = C^k_n*p^k*(1p)^{nk}= C^4_6*(\frac{1}{2})^4*(1\frac{1}{2})^{64}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
Consider this:
We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).
Answer: C.
_________________
 If you think you can,you can If you think you can't,you are right.



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 10:17
mojorising800 wrote: hi bunuel could u plz provide a few application qs. to this formula? TIA Bunuel wrote: NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1p)^{nk}\)
\(P = C^k_n*p^k*(1p)^{nk}= C^4_6*(\frac{1}{2})^4*(1\frac{1}{2})^{64}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
Consider this:
We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).
Answer: C. Check the Probability and Combinatorics chapters in Math Book (link in my signature): theory, examples and links to the problems.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 25 Dec 2009
Posts: 78

Re: probability Qs.. attention
[#permalink]
Show Tags
Updated on: 09 Jan 2010, 15:29
We missed this in this thread : Question: If we have n objects out of which p are alike, q are alike, and we are to choose r objects out of it, how many possible number of comnitations are possible ? A permutation problem, answer is nPr/p!*q! as in the discussion above , but in case of combination what is formula? ?? Lets say we have 10 books, out of which 3 are copies of same book, 4 are copies of another book in how many ways 4 books can be selected ? so answer is : 5C4 I beleive i.e. 1034 are different books i.e. 3 different books, 3 books of one type can be treated as one, 4 other different books can be treated as one again.. so total 5 books , out of which 4 can be selected. @Bunel : While thinking of a question for you, I got the answer too . THANKS. Question to Bunel : I have flexed my muscles in number roperties but need to sweat more, could you refer me some part over this site for basic and advance review of number properties, as well as well "hard but GOOD questions" .
Originally posted by GMATMadeeasy on 09 Jan 2010, 13:59.
Last edited by GMATMadeeasy on 09 Jan 2010, 15:29, edited 1 time in total.



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 14:15



Manager
Joined: 27 Apr 2008
Posts: 173

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 14:53
Ah I see where I got it wrong now. Thanks for the Math book reference.



Manager
Joined: 27 Apr 2008
Posts: 173

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 14:57
Bunuel wrote: NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1p)^{nk}\)
\(P = C^k_n*p^k*(1p)^{nk}= C^4_6*(\frac{1}{2})^4*(1\frac{1}{2})^{64}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
By the way, the formula should be this instead: \(P = C^n_k*p^k*(1p)^{nk}\)



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: probability Qs.. attention
[#permalink]
Show Tags
09 Jan 2010, 15:08
mrblack wrote: Bunuel wrote: NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1p)^{nk}\)
\(P = C^k_n*p^k*(1p)^{nk}= C^4_6*(\frac{1}{2})^4*(1\frac{1}{2})^{64}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
By the way, the formula should be this instead: \(P = C^n_k*p^k*(1p)^{nk}\) It's the same. Since n>=k, you can write as \(nCk\), \(C(n,k)\), \(C(k,n)\), \(C^n_k\), \(C^k_n\), it's clear what is meant. Actually in different books you can find different forms of writing this. Walker in his topic used \(C^n_k\), but \(C^k_n\) is also correct.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 13 Jan 2015
Posts: 11

The probability that a family with 6 children has exactly
[#permalink]
Show Tags
Updated on: 21 Jan 2017, 04:19
Bunuel can you please explain the logic behind 'p' being 1/2
Originally posted by ram186 on 21 Jan 2017, 04:16.
Last edited by ram186 on 21 Jan 2017, 04:19, edited 1 time in total.



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: The probability that a family with 6 children has exactly
[#permalink]
Show Tags
21 Jan 2017, 04:18
ram186 wrote: Bunuel wrote: The probability that a family with 6 children has exactly four boys is:
A. 1/3 B. 1/64 C. 15/64 D. 3/8 E. none of the above
NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1p)^{nk}\)
\(P = C^k_n*p^k*(1p)^{nk}= C^4_6*(\frac{1}{2})^4*(1\frac{1}{2})^{64}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
Consider this:
We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).
Answer: C.[/Bunuel can you please explain the logic behind 'p' being 1/2] The probability of having a girl = the probability of having a boy = 1/2.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 11 Jun 2015
Posts: 19
Location: Iran (Islamic Republic of)
Concentration: Accounting, Finance
WE: Education (Education)

Re: The probability that a family with 6 children has exactly
[#permalink]
Show Tags
26 Aug 2018, 00:09
How we can infer that the chance of having a boy is 0.5? What if it were 0.2 ? Don't you think the question stem must at least asserts that it is a normal family with equal change of getting a boy or a girl? Correct me, if my thoughts is wrong. Thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: The probability that a family with 6 children has exactly
[#permalink]
Show Tags
26 Aug 2018, 03:15



Manager
Joined: 03 Feb 2018
Posts: 75

Re: The probability that a family with 6 children has exactly
[#permalink]
Show Tags
09 Sep 2018, 04:16
GMATMadeeasy wrote: The probability that a family with 6 children has exactly four boys is:
A. 1/3 B. 1/64 C. 15/64 D. 3/8 E. none of the above Hi, Please tell me what is the meaning of this question. I am not able to get any relation between a chicken and boy.



Math Expert
Joined: 02 Sep 2009
Posts: 51119

Re: The probability that a family with 6 children has exactly
[#permalink]
Show Tags
09 Sep 2018, 06:18




Re: The probability that a family with 6 children has exactly &nbs
[#permalink]
09 Sep 2018, 06:18



Go to page
1 2
Next
[ 22 posts ]



