In a Family With 6 Children Excluding Multiple Births What Is the Probability of Having 6 Girls

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The probability that a family with 6 children has exactly [#permalink]   Updated on: 10 February 2012, xi:42

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The probability that a family with six children has exactly four boys is:

A. 1/3
B. ane/64
C. fifteen/64
D. 3/eight
E. none of the higher up

Originally posted by GMATMadeeasy on 08 Jan 2010, 16:56.
Last edited by Bunuel on 10 Feb 2012, 11:42, edited 1 fourth dimension in total.

Edited the question and added the OA

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Re: probability Qs.. attending [#permalink]   08 January 2010, 22:35

The probability that a family with six children has exactly four boys is:

A. 1/3
B. one/64
C. 15/64
D. iii/eight
E. none of the above

NOTE: If the probability of a sure event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of information technology occurring \(chiliad\) times (iv times in our instance) in \(n\)-time (half dozen in our case) sequence is:

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

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

Consider this:

We are looking for the example BBBBGG, probability of each B or One thousand is \(\frac{1}{ii}\), hence \(\frac{1}{2^half dozen}\). Only BBBBGG can occur (these letters can be arranged) in \(\frac{half dozen!}{4!2!}\) # of ways. Then, the full probability would exist \(\frac{6!}{four!2!}*\frac{1}{2^6}=\frac{15}{64}\).

Answer: C.
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Re: probability Qs.. attention [#permalink]   09 January 2010, 03:eleven

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

And then there are vi combinations here. But since BG can also be GB, we have to multiply past 2. Nosotros have to multiply by another 2 because there are 2 As. Therefore, in that location are 6x2x2=24 combinations.

We besides take to include the instances where GG occurs at the end or the forepart, that is:
GGBBBB
BBBBGG
(all the other instances such as BBGGBB accept been taken care of in the calculation at the top)

And so we have 24+2=26 possibilities of iv boys and 2 girls.

The total number of possibilities is two^6=64. Therefore, the probability is 26/64, which is thirteen/32. The reply is E, none of the above.

Your solution can be simplified this way as well:

Full number of events : 2^six every bit each position tin be filled in exactly two ways
Events in favor : (four Boys and Two Girls; Information technology is like six objects in which 4 are alike and 2 others are alike) so half-dozen!/2!4!

If information technology were a probability question, I will follow Bunuel'south aproach merely the solution above is also of import if it is counting problem.

Question: If we have due north 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 northward!/p!*q! as in the question above

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Re: probability Qs.. attention [#permalink]   09 Jan 2010, 04:18

I think this way:

For each child we've 2 options i.eastward., that child tin either be a boy or a girl.Hence we accept 2^6 = 64 total outcomes.
At present out of six children whatever 4 of them can boys and this can exist washed in 6C4 = 15 ways.
Now here I don't recollect that ordering does thing in this case.

And then the probability comes out to be 15/64.

P.South. Delight 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]   09 Jan 2010, 04:56

ashueureka wrote:

I think this fashion:

For each child we've two options i.east., that child tin can either be a boy or a girl.Hence nosotros accept two^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 hither I don't think that ordering does matter in this case.

Then 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 fashion of solving is also correct. You've already considered all possible arrangements for BBBBGG with 6C4, which is 6!/4!two!. If you lot look at my solution and at yours you'll run across that both wrote the same formulas but with dissimilar approach.
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Re: probability Qs.. attention [#permalink]   09 January 2010, 06:07

chetan2u wrote:

how-do-you-do bunuel , this is binomial distr(seen a few q on gmat).... exercise we accept q in gmat on which we require hypergeometric distr

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

This can exist considered every bit hypergeometric distribution, simply it's quite simple: 3C2*5C3/8C5.
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Re: probability Qs.. attending [#permalink]   09 Jan 2010, 07:06

how-do-you-do bunuel
could u plz provide a few application qs. to this formula?
TIA

Bunuel wrote:

Annotation: If the probability of a sure consequence is \(p\) (\(\frac{i}{2}\) in our example), then the probability of it occurring \(k\) times (4 times in our case) in \(northward\)-time (6 in our case) sequence is:

\(P = C^k_n*p^k*(1-p)^{due north-1000}\)

\(P = C^k_n*p^k*(1-p)^{northward-k}= C^4_6*(\frac{one}{2})^four*(1-\frac{1}{2})^{vi-4}= C^4_6*(\frac{1}{2})^4*(\frac{one}{two})^{2}=\frac{half-dozen!}{4!2!}*\frac{ane}{2^6}=\frac{15}{64}\)

Consider this:

We are looking for the example BBBBGG, probability of each B or 1000 is \(\frac{1}{2}\), hence \(\frac{i}{2^six}\). Only BBBBGG can occur (these letters can be arranged) in \(\frac{vi!}{4!2!}\) # of ways. So, the total probability would be \(\frac{half dozen!}{4!ii!}*\frac{ane}{ii^half-dozen}=\frac{xv}{64}\).

Answer: C.

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Re: probability Qs.. attention [#permalink]   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 \(p\) (\(\frac{one}{2}\) in our example), then the probability of information technology occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

\(P = C^k_n*p^yard*(i-p)^{n-chiliad}\)

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

Consider this:

We are looking for the example BBBBGG, probability of each B or M is \(\frac{1}{two}\), hence \(\frac{1}{two^six}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{vi!}{iv!2!}\) # of ways. Then, the total probability would be \(\frac{six!}{4!2!}*\frac{ane}{2^six}=\frac{15}{64}\).

Respond: C.

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Re: probability Qs.. attending [#permalink]   09 Jan 2010, 14:57

Bunuel wrote:

NOTE: If the probability of a certain event is \(p\) (\(\frac{i}{ii}\) in our case), then the probability of information technology occurring \(1000\) times (iv times in our case) in \(n\)-time (6 in our case) sequence is:

\(P = C^k_n*p^1000*(1-p)^{n-k}\)

\(P = C^k_n*p^k*(ane-p)^{n-k}= C^4_6*(\frac{i}{2})^4*(1-\frac{1}{2})^{six-4}= C^4_6*(\frac{i}{two})^4*(\frac{1}{2})^{2}=\frac{half-dozen!}{4!2!}*\frac{one}{two^half-dozen}=\frac{15}{64}\)

By the mode, the formula should be this instead:

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

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Re: probability Qs.. attention [#permalink]   09 Jan 2010, xv:08

mrblack wrote:

Bunuel wrote:

NOTE: If the probability of a sure effect is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(grand\) times (4 times in our case) in \(n\)-time (6 in our example) sequence is:

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

\(P = C^k_n*p^k*(one-p)^{n-k}= C^4_6*(\frac{1}{ii})^4*(1-\frac{ane}{2})^{6-iv}= C^4_6*(\frac{i}{two})^4*(\frac{1}{2})^{2}=\frac{6!}{four!2!}*\frac{1}{2^6}=\frac{15}{64}\)

By the way, the formula should be this instead:

\(P = C^n_k*p^1000*(one-p)^{n-chiliad}\)

It's the same. Since n>=chiliad, you lot can write every bit \(nCk\), \(C(n,k)\), \(C(1000,northward)\), \(C^n_k\), \(C^k_n\), it'south clear what is meant. Really in dissimilar books you can discover different forms of writing this. Walker in his topic used \(C^n_k\), but \(C^k_n\) is also right.
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The probability that a family with 6 children has exactly [#permalink]   Updated on: 21 Jan 2017, 04:nineteen

Bunuel tin can you lot please explicate the logic backside 'p' being 1/2

Originally posted by ram186 on 21 Jan 2017, 04:16.
Last edited past ram186 on 21 Jan 2017, 04:19, edited one fourth dimension in full.

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Re: The probability that a family with vi children has exactly [#permalink]   21 Jan 2017, 04:18

ram186 wrote:

Bunuel wrote:

The probability that a family with 6 children has exactly iv boys is:

A. i/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above

Note: If the probability of a certain consequence is \(p\) (\(\frac{1}{ii}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(due north\)-fourth dimension (half-dozen in our case) sequence is:

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

\(P = C^k_n*p^k*(1-p)^{north-g}= C^4_6*(\frac{ane}{ii})^4*(1-\frac{i}{ii})^{half dozen-4}= C^4_6*(\frac{one}{2})^4*(\frac{1}{2})^{2}=\frac{half dozen!}{4!two!}*\frac{1}{ii^6}=\frac{15}{64}\)

Consider this:

We are looking for the example BBBBGG, probability of each B or K is \(\frac{1}{2}\), hence \(\frac{1}{ii^half-dozen}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{iv!2!}\) # of ways. And so, the total probability would be \(\frac{six!}{4!2!}*\frac{i}{2^6}=\frac{15}{64}\).

Answer: C.[/Bunuel tin can y'all please explain the logic behind 'p' beingness i/two]

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]   26 Aug 2018, 00:09

How nosotros can infer that the chance of having a boy is 0.five?
What if it were 0.2 ?
Don't you think the question stem must at to the lowest degree asserts that information technology is a normal family with equal change of getting a boy or a girl?
Correct me, if my thoughts is incorrect.
Cheers.

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

soleimanian wrote:

How we can infer that the take a chance of having a boy is 0.5?
What if it were 0.2 ?
Don't you lot recall the question stalk must at to the lowest degree asserts that it is a normal family with equal alter of getting a boy or a girl?
Right me, if my thoughts is incorrect.
Thanks.

Proper GMAT question would specify this. Then, don't worry on the actual examination everything will be unambiguous.
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Re: The probability that a family with 6 children has exactly [#permalink]   05 Nov 2020, 00:02

P(Boy) = \(\frac{i}{two}\)

Total: 6 and we need '4' boys

=> \(^6{C_4} * (\frac{ane}{2})^4 * (\frac{one}{two})^2\)

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

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

Answer C
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Re: The probability that a family unit with 6 children has exactly [#permalink]   thirteen Dec 2020, 05:29

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. iii/8
E. none of the above

Solution:

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

½ ten ½ ten ½ ten ½ 10 ½ ten ½ = 1/64

Nevertheless, since iv boys and two girls tin be arranged in six!/(4! ten 2!) = (6 x v)/2 = xv means, the probability of having iv boys and 2 girls (in whatsoever social club) is 15 x 1/64 = fifteen/64.

Reply: C

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Re: The probability that a family unit with 6 children has exactly [#permalink]   12 Feb 2021, 18:37

Bunuel wrote:

The probability that a family with 6 children has exactly 4 boys is:

A. 1/three
B. ane/64
C. fifteen/64
D. iii/8
Due east. none of the higher up

Annotation: If the probability of a certain upshot is \(p\) (\(\frac{i}{2}\) in our case), and so the probability of it occurring \(m\) times (4 times in our instance) in \(n\)-time (six in our case) sequence is:

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

\(P = C^k_n*p^g*(1-p)^{n-yard}= C^4_6*(\frac{one}{2})^four*(1-\frac{1}{two})^{6-iv}= C^4_6*(\frac{one}{2})^4*(\frac{one}{2})^{2}=\frac{6!}{4!two!}*\frac{1}{ii^vi}=\frac{fifteen}{64}\)

Consider this:

We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{i}{two^half dozen}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the full probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).

Reply: C.

Bunuel I get the math, simply why should nosotros care about the social club? That'southward 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]   xiii Feb 2021, 01:49

CEdward wrote:

Bunuel wrote:

The probability that a family with 6 children has exactly four boys is:

A. one/3
B. i/64
C. 15/64
D. iii/8
E. none of the in a higher place

Annotation: If the probability of a certain outcome is \(p\) (\(\frac{1}{2}\) in our example), so the probability of it occurring \(k\) times (iv times in our case) in \(north\)-fourth dimension (six in our case) sequence is:

\(P = C^k_n*p^k*(ane-p)^{north-k}\)

\(P = C^k_n*p^grand*(one-p)^{n-m}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{half dozen-4}= C^4_6*(\frac{1}{ii})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{two^6}=\frac{15}{64}\)

Consider this:

We are looking for the instance BBBBGG, probability of each B or M is \(\frac{i}{ii}\), hence \(\frac{1}{two^vi}\). But BBBBGG tin can occur (these messages tin can be arranged) in \(\frac{6!}{iv!2!}\) # of ways. And so, the total probability would be \(\frac{6!}{iv!2!}*\frac{1}{2^vi}=\frac{fifteen}{64}\).

Answer: C.

Bunuel I become the math, but why should we care near the order? That'due south the part that doesn't make sense to me. Isn't it sufficient to only accept 4 boys and 2 girls, irrespective of the club?

A family tin have iv boys and two girls in fifteen different ways (BBBBGG, GGBBBB, GBBBBG, GBBGBB, ...). Each of them has the probability of 1/two^6, then the overall probability is the sum of these 15 different cases, giving the final answer of xv*1/2^6.
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Re: The probability that a family with half-dozen children has exactly [#permalink]   13 Jul 2021, 06:16

the question should be edited and exist 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]   xix Jul 2021, 08:26

Bunuel Hi, is this question GMAT centric? I found it very vague

Re: The probability that a family unit with 6 children has exactly [#permalink]

nineteen Jul 2021, 08:26

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