The probability that a family with 6 children has exactly

Question

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

Bunuel · Accepted Answer

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.

GMATMadeeasy · Answer

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

ashueureka · Answer

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.

Bunuel · Answer

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.

Bunuel · Answer

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.

mojorising800 · Answer

hi bunuel
could u plz provide a few application qs. to this formula?
TIA

Bunuel · Answer

Check the Probability and Combinatorics chapters in Math Book (link in my signature): theory, examples and links to the problems.

mrblack · Answer

By the way, the formula should be this instead:

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

Bunuel · Answer

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.

ram186 · Answer

Bunuel can you please explain the logic behind 'p' being 1/2

Bunuel · Answer

The probability of having a girl = the probability of having a boy = 1/2.

soleimanian · Answer

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.

Bunuel · Answer

Proper GMAT question would specify this. So, don't worry on the actual exam everything will be unambiguous.

MathRevolution · Answer

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

ScottTargetTestPrep · Answer

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

CEdward · Answer

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?

Bunuel · Answer

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.

naveeng15 · Answer

Its a binomial situation ,

since the P(having Boy or girl ) is independent and its 1/2 we can use the formula

P(E) = nCr P(success)^r P(failure)^(n-r)

= 6C4 (1/2)^4 (1/2)^2
= 15/64

Mugdho · Answer

KarishmaB  
 Bunuel 
 chetan2u

GMATinsight

why does it have to be a case of ordering? question didn’t say anything about order...it simply asked if there are 4 boys and 2 girls...

if we consider the cases, as-

6 Girls
5 girls and 1 boy
4 girls and 2 boys
3 girls and 3 boys
2 girls and 4 boys
1 girl and 5 boys

So there are 7 cases...and 4 boys can happen only in one way...

so probability : 1/7

what's wrong with this reasoning?

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