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The probability that in a family there are $n$ children is $ap^n$ $a\gt 0,\;0\lt p\lt 1$ And the probability of a boy is equal to that of a girl in the family. What is the probability that there are exactly $k(\gt 1)$boys in the family And given that a family has at least $1$ boy what is the probability that there are two or more children in the family. I found the answer to first part as $$\frac {(1-p)p^k2^{(1-k)}}{2-p}$$ I just calculated this by starting with one boy then two boys.....and got this. My question is is it correct and if not please tell the correct method And for the 2nd part i dont have any clue how to proceed For the first part i started with the probability of having $1$ boy,that can happen as (1child 1 boy) or (2children,1 boy) or (3children 1 boy) and so on $P(boy=1)=P(1 boy| 1 child)P(1 child) +P(1 boy|2children)P(2children)+........ $ $=\frac {ap}{2}+\frac {ap^2}{2^2}+\frac {ap^3}{p^3}....$ $=\frac {ap}{2-p}=\frac {(1-p)p}{2-p}$

$P(2 boys)=P(2boys|2children)P(2 children)+P(2 boys|3children)P(3 children)....$

which as above after simplyfying i got $=\frac {(ap^2)2^2}{2(2-p)}$

using this i got $$P(k boys)=\frac {(1-p)p^k2^{(1-k)}}{2-p}$$

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  • $\begingroup$ @zhoraster: I rolled back your edit because it changed the meaning of the question. There was no condition $n\ge1$. $\endgroup$
    – joriki
    Mar 17 '16 at 9:39
  • $\begingroup$ @zhoraster: Independence of $a$ is correct, since $a$ is fixed by normalization to be $1-p$. $\endgroup$
    – joriki
    Mar 17 '16 at 9:40
  • $\begingroup$ @joriki, it is meaningless in this form, because of some negative probabilities. I think that $P(n\text{ children}) = ap^n$, $n\ge 1$, and $P(0\text{ children}) = 1-ap/(1-p)$. (This way it is formulated in our problem book :) probability.univ.kiev.ua/userfiles/kmv/gkr-problems.pdf, p. 30, problem 10.) $\endgroup$
    – zhoraster
    Mar 17 '16 at 9:41
  • $\begingroup$ @zhoraster: Could be, but I don't think one should change someone else's question based on a guess what they might have meant -- there's always the possibility to ask for clarifications. $\endgroup$
    – joriki
    Mar 17 '16 at 9:43
  • $\begingroup$ @Upstart: $a\lt0$ makes no sense; that would lead to negative probabilities. $\endgroup$
    – joriki
    Mar 17 '16 at 9:43
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By normalization, $a=1-p$, so the probability for $n$ children is $(1-p)p^n$, a geometric distribution that we can interpret as if before each new child the parent/s throw/s a coin with success probability $p$ whether to have another child.

Assuming binary gender (as you seem to be doing), the probability of each child being a boy is $\frac12$. Then, considering boys only, the probability $x$ to stop having boys is $1-p +p\cdot\frac12\cdot x$, since we can stop either immediately or after having a girl. Solving for $x$ yields $x=(1-p)/(1-\frac p2)$. Thus the probability for $k$ boys in the family is

$$ \frac{1-p}{1-\frac p2}\left(1-\frac{1-p}{1-\frac p2}\right)^k=\frac{1-p}{1-\frac p2}\left(\frac p{2-p}\right)^k\;. $$

Given that the family has at least one boy, it has two or more children unless it has exactly that one boy and no other children, and the probability for that is $p\cdot\frac12\cdot(1-p)=\frac12p(1-p)$, so the desired probability is $1-\frac12p(1-p)$.

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  • $\begingroup$ I dont understand joriki..what is $x$. $\endgroup$
    – Upstart
    Mar 17 '16 at 11:00
  • $\begingroup$ @Upstart: I defined it: "the probability $x$ to stop having boys". $\endgroup$
    – joriki
    Mar 17 '16 at 11:03

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