\begin{align} P_k&= \text{probability of a randomly chosen family having exactly k children} \\ &= αp_k,\qquad k=1,2,.. \end{align} $$$$

Suppose that all sex distribution of $k$ children are equally likely. Find the probability that a family has exactly $r$ boys, $r≥1$. Find the conditional probability that a family has at least two boys, given that it has at least one boy.

So $k=\text{no. of children}$ and $r=\text{no. of boys}$

If I were to get the probability that a family has exactly $r$ boys then that is

$$P(R=r)=\sum P(K=k)P(R=r|K=k)$$

I substituted values to get

$$P(R=r)=\sum αp_k(_kC_r(0.5)^k)$$

Now the probability being asked is

$$P(R≥2 | R≥1)= P(R≥2 ∩ R≥1) / P(R≥1)$$ and $(R≥2 ∩ R≥1)$ is just $R≥2$ so

$$P(R≥2 | R≥1)= P(R≥2) / P(R≥1)$$



If you want a simpler expression start simplifying earlier. I will denote $q_i = P(R=i)$. At the point where you have:

$$ \frac{1 - q_0 - q_1}{1 - q_0}, $$

You can break apart the numerator and get:

$$ 1 - \frac{q_1}{1 - q_0}. $$


Alright since that didn't do the trick we'll have to get our hands dirty. First lets shift our focus to exact strings of children. So instead of a binomial we'll have lots of independent Bernoulli trials. Consider the string:

ggbggbggg = (ggbgg)(b)(ggg).

Letting $q=0.5$, we can factor the probability of this string into:

$$ ap^9q^9 = (ap^5q^5)(pq)(p^3q^3) $$

Every string containing exactly 2 boys will look like this, a string containing exactly 1 boy, another boy, a (possibly empty) string containing containing only girls. We can find the probability of all two boy strings starting with (ggbgg) by multiplying $ (ap^5q^5)$ by

$$ \sum_{i=1}^{\infty} (pq)^i. $$

Similarly we can find the probability of all strings containing two boys by taking

$$ P(R=1) \cdot \sum_{i=1}^{\infty} (pq)^i. $$

I'll let you take it from here (geometric series simplify nicely)

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  • $\begingroup$ Thank you for the reply! I tried this method but I ended up with an expression that doesn't seem in its simplest form. And I still haven't used P0. $\endgroup$ – Rinrin Sep 13 '15 at 2:27
  • $\begingroup$ @Rinrin did you ever end up seeing this? $\endgroup$ – jlimahaverford Sep 26 '15 at 13:59

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