The probability that a family with 6 children has exactly

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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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

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

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

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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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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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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the question should be edited and be told that the Probability of Boy or a girl is 1/2.

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