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The probability that a family with 6 children has exactly

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GMATMadeeasy
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The probability that a family with 6 children has exactly [#permalink] New post   Updated on: 10 Feb 2012, 12:42
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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
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C

Originally posted by GMATMadeeasy on 08 Jan 2010, 17:56.
Last edited by Bunuel on 10 Feb 2012, 12:42, edited 1 time in total.
Edited the question and added the OA
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Re: probability Qs.. attention [#permalink] New post   08 Jan 2010, 23:35
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 04:11
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 05:18
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 05:56
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 07:07
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 08:06
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 11:17
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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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 15:57
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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Re: probability Qs.. attention [#permalink] New post   09 Jan 2010, 16:08
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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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The probability that a family with 6 children has exactly [#permalink] New post   Updated on: 21 Jan 2017, 05:19
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Bunuel can you please explain the logic behind 'p' being 1/2

Originally posted by ram186 on 21 Jan 2017, 05:16.
Last edited by ram186 on 21 Jan 2017, 05:19, edited 1 time in total.
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Re: The probability that a family with 6 children has exactly [#permalink] New post   21 Jan 2017, 05:18
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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:

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

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


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.[/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.
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Re: The probability that a family with 6 children has exactly [#permalink] New post   26 Aug 2018, 01:09
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.
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Re: The probability that a family with 6 children has exactly [#permalink] New post   26 Aug 2018, 04:15
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soleimanian wrote:
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.


Proper GMAT question would specify this. So, don't worry on the actual exam everything will be unambiguous.
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Re: The probability that a family with 6 children has exactly [#permalink] New post   05 Nov 2020, 01:02
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P(Boy) = \(\frac{1}{2}\)

Total: 6 and we need '4' boys

=> \(^6{C_4} * (\frac{1}{2})^4 * (\frac{1}{2})^2\)

=> 15 * (\(\frac{1}{64}\))

=> \(\frac{15}{64}\)

Answer C
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Re: The probability that a family with 6 children has exactly [#permalink] New post   13 Dec 2020, 06:29
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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

Solution:

Letting B represent a boy child and G represent a girl, the probability of 4 boys and 2 girls, in the specific order of BBBBGG, is:

½ x ½ x ½ x ½ x ½ x ½ = 1/64

However, since 4 boys and 2 girls can be arranged in 6!/(4! x 2!) = (6 x 5)/2 = 15 ways, the probability of having 4 boys and 2 girls (in any order) is 15 x 1/64 = 15/64.

Answer: C
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Re: The probability that a family with 6 children has exactly [#permalink] New post   12 Feb 2021, 19:37
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:

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

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


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.


Bunuel I get the math, but why should we care about the order? That's the part that doesn't make sense to me. Isn't it sufficient to just have 4 boys and 2 girls, irrespective of the order?
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Re: The probability that a family with 6 children has exactly [#permalink] New post   13 Feb 2021, 02:49
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CEdward 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:

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

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


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.


Bunuel I get the math, but why should we care about the order? That's the part that doesn't make sense to me. Isn't it sufficient to just have 4 boys and 2 girls, irrespective of the order?


A family can have 4 boys and 2 girls in 15 different ways (BBBBGG, GGBBBB, GBBBBG, GBBGBB, ...). Each of them has the probability of 1/2^6, so the overall probability is the sum of these 15 different cases, giving the final answer of 15*1/2^6.
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Re: The probability that a family with 6 children has exactly [#permalink] New post   13 Jul 2021, 07:16
the question should be edited and be told that the Probability of Boy or a girl is 1/2.
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Re: The probability that a family with 6 children has exactly [#permalink] New post   19 Jul 2021, 09:26
Bunuel Hi, is this question GMAT centric? I found it very vague
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Re: The probability that a family with 6 children has exactly [#permalink]
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