0
$\begingroup$

\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

$$P(R≥2 | R≥1)= P(R≥2) / P(R≥1)$$

????

$\endgroup$
1
$\begingroup$

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:

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

edit

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)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.