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The probability that a family with 6 children has exactly

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Feb 2012, 11:42, edited 1 time in total.
Edited the question and added the OA
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New post 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.
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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:

\(P = C^k_n*p^k*(1-p)^{n-k}\)

\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)


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.
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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
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New post 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.
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New post 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.
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New post 09 Jan 2010, 06:07
chetan2u wrote:
hi bunuel , this is binomial distr(seen a few q on gmat).... do we have q in gmat on which we require hypergeometric distr


Simple ones. For example: there are 3 men and 5 women, what's the probability of choosing 5 people out of which 2 are men?

This can be considered as hypergeometric distribution, but it's quite simple: 3C2*5C3/8C5.
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New post 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:

\(P = C^k_n*p^k*(1-p)^{n-k}\)

\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)


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.

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New post 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:

\(P = C^k_n*p^k*(1-p)^{n-k}\)

\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)


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.
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New post 09 Jan 2010, 13:59
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. 10-3-4 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. :-D

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" .

Last edited by GMATMadeeasy on 09 Jan 2010, 15:29, edited 1 time in total.
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New post 09 Jan 2010, 14:15
GMATMadeeasy wrote:
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 nCr/p!*q! as in the discussion above , but in case of combination what is formula? ??


Can you please give an example of this, to understand correctly your question?
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New post 09 Jan 2010, 14:53
Ah I see where I got it wrong now. Thanks for the Math book reference.
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New post 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:

\(P = C^k_n*p^k*(1-p)^{n-k}\)

\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)



By the way, the formula should be this instead:

\(P = C^n_k*p^k*(1-p)^{n-k}\)
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New post 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:

\(P = C^k_n*p^k*(1-p)^{n-k}\)

\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)



By the way, the formula should be this instead:

\(P = C^n_k*p^k*(1-p)^{n-k}\)


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.
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