what is the probability of a couple who has four girls and is trying again for a boy. what is the probability that the next kid will be a girl? A couple has four kids already, all girls. the couple would like to have a son and would like to give it another try. what is the probability that the next kid will be a girl?
 A: If we presume that we know nothing about the actual probability of having a girl, $p$, then we can use Hinkley's predictive likelihood for the binomial to predict the probability that the next birth will also be a girl, assuming births are conditionally independent, given $p$. This is sensible, since the probability that a couple will have a girl or boy is affected primarily by the X/Y ratio of the male gametes, and to some extent selective factors in the female. Therefore, while overall the bias is slightly toward male births, there will be overdispersion around this mean due to individual propensities that are at odds with the overall average.
Let $g$ be the number of girls born to date and $n$ be the number of children born to-date. For general prediction, we want to know the probability that there will be $r$ girls in the next $s\geq r$ births.
$P(r|g,n,s) = K\frac{{n \choose g}{s \choose r}}{n+s\choose g+r}$, where $K$ is a normalizing constant to make $\sum\limits_{r=0}^s\frac{{n \choose g}{s \choose r}}{n+s\choose g+r} = 1$
In your case, we are dealing with $s=1$ and $n=g=4$, which greatly simplifies things, since  there are only two outcomes ($r=0,r=1$) and ${1\choose 0} = {1 \choose 1} = 1$. 
The simplified formula is:
$P(r|4,4,1) = K\frac{{4 \choose 4}{1 \choose r}}{5\choose 4+r}= K\frac{(4+r)!(r)!}{120} = K\left( \frac{1}{5}+\frac{4r}{5}\right)$, (since $r$ can only be $1$ or $0$)
In this case, there are only two values of $r$ so getting $K$ is easy:
$P(r=0|4,4,1) + P(r=1|4,4,1) = 1 \rightarrow K \left( \frac{1}{5} + 1\right) = 1 \therefore K=\frac{5}{6}$
Given the above, the probability that the next child is a girl, given that all 4 previous children were girls, is: $P(r=1|4,4,1)=\frac{5}{6} \approx 83\%$
A more detailed exposition using more sophisticated models can be found here - it studies possible models for the overdispersion of actual sex ratios in the human population, which lends evidence that the probability of having a girl can vary either between families or even within a single family over time.
A: i agree. having the 4 girls in the beginning means nothing: it does not affect if the next child will be a girl or a boy. So it is independent from everything and the chances is 50%
A: This problem seems to be homework, so I assume the problem is not a complex biology problem and there's a 50/50 chance of boy or girl. We can assume the probabilities are independent to each other, so the probability is $50\%$.
Notes: The average male/female ratio in the world is about $1:1$, and it's safe to assume half of sperm contain an X chromosome and other half contain Y from meiosis. I think it's also safe to assume fertilization is independent.
