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Consider families with three children. What is the probability that all are boys given that:

(i) the first is a boy (ii) there is at least one boy

I don't really understand this question. Assuming that the likelihood of having a male child is the same of any of having a female child

P(male) = 0.5

Would my answer for (i) be (0.5)(0.5)(0.5) = 0.125

I don't think the question really makes sense as I am told that all the children are male.

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  • $\begingroup$ "as I am told that all the children are male" Really? Where are you told this? $\endgroup$ – Did Aug 2 '14 at 14:35
  • $\begingroup$ "What is the probability that all are boys given that (i) the first is a boy and (ii) at least one is a boy. Does this not mean that I am working with a sample space of exclusively male children? $\endgroup$ – Roy Sheehan Aug 2 '14 at 14:58
  • $\begingroup$ No. $ $ $ $ $ $ $\endgroup$ – Did Aug 2 '14 at 18:14
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Note that order matters in this case (since there can be multiple orders for any one configurations) and we have:

1 configuration with 3 boys

3 configurations with 2 boys and 1 girl

3 configurations with 2 girls and 1 boy

1 configuration with 3 girls

all configs with equal probability of 1/8 (meaning 3/8 for any possibility with 3 configurations).

With i) as your hint, we eliminate the last configuration, one of the three with 2 boys and 1 girl (i.e. girl-boy-boy) and two of the three with 2 girls and 1 boy (i.e. all except boy-girl-girl). Thus we're left with 4 and all boys is one of the 4, so the chance is 1/4.

If ii) is your hint, then we only eliminate the last configuration (since there's at least 1 boy). Then there's 7 configurations left, with 3 boys being one of the seven, thus the probability is 1/7.

EDIT: Let's consider the case of $n$ kids, with either of these hints, and we assess the probability of $n$ boys.

i) With the first given as a boy, look at the next $n-1$ kids. The chance of all of them being a boy is $(\frac{1}{2})^{n-1}$.

ii) Normally there are $2^n$ configurations, each with equal probability. Given that there's at least one boy, we eliminate the possibility of $n$ girls so there's $2^n - 1$ possibilities. With $n$ boys being one of these possibilities, the probability is $\frac{1}{2^n-1}$.

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  • $\begingroup$ Is there a way I can find out the number of configurations for other questions? As in is there a formula I can use for similar situations, or have you just been able to look at the question and see the answer from experience? $\endgroup$ – Roy Sheehan Aug 2 '14 at 14:43
  • $\begingroup$ Well I don't know what you mean by "similar", but I edited the solution to getting $n$ boys out of $n$ kids with these same hints. $\endgroup$ – Bridgeburners Aug 2 '14 at 15:01
  • $\begingroup$ My hero! Thanks a lot for all your help. $\endgroup$ – Roy Sheehan Aug 2 '14 at 15:08
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Hints:

If $X$ denotes the number of boys among the two children that are not the first then in (i) you are asked to find $P\left\{ X=2\right\} $.

If $B$ denotes the number of boys among all three children then in (ii) you are asked to find $P\left\{ B=3\mid B\geq1\right\} $.

Note that in both cases you are dealing with a binomial distribution.

You can also look arrangements like BBB, BGB, ... et cetera and count.

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Well, for (i):

Since you already know that the first child is a boy, you just need to count the cases for the other two children; All the possible cases are: BBB (the only favorable case), BBG, BGB, BGG, which gives you $1/4$.

And (ii):

So, since all of the possible cases are BBB, BBG, BGB, ... (there are 8 of them), and we know that there is at least one boy, we're left with $1/7$.

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  • $\begingroup$ I'm sure that there is some kind of a biological, natural distribution for these things, but I'm also pretty sure it's quite complex and probably should be disregarded here :) $\endgroup$ – Eutherpy Aug 2 '14 at 14:37
  • $\begingroup$ @Eurthepy Any idea what $$(ii)$$ would look like? $\endgroup$ – Roy Sheehan Aug 2 '14 at 14:40
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for (ii), the complementary is to have only girls.

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