A married couple agreed to continue bearing a new child until they get two boys, but not more than $5$ children. Assuming that each time that a child is born the probability that it is a boy is $0.5$ independent from all other times, find the probability that the couple has at least two girls.
[Math] Probability by bearing of children
probability
Related Solutions
For part I:
Order the children from youngest to oldest. A priori there are $2^5=32$ equally probable ways to assign gender to the collection. We exclude "all girls", making for $31$ cases. Other than $\{G,G,G,G,B\}$ (a $\frac 1{31}$ case), every combination with a boy in the last spot can be paired with one with a girl in the last slot. Thus the probability that the eldest is a boy is $$\frac 12\times \frac {30}{31}+\frac 1{31}=\frac {16}{31}$$
For part II:
your first line already specifies two boys and three girls, you don't need the second factor.
For part III:
There are $10$ equally probable scenarios in which there are $2$ boys and $3$ girls. $4$ of these have the eldest child a boy, so $\frac 4{10}$.
Have a look and count the number of succesful outcomes compared to the total number:
If you correct for biological factors, however, the answer may be skewed in either direction. Perhaps boys are a little more likely than girls, so I have heard.
If we only dealt with gender, not the day of week, your figure would have been correct since the table would then simply be:
The two things that confuse our intuitions are:
It is counter intuitive that someone would NOT mention which kid (the oldest, the youngest) which allows for the overlapping areas in the two tables.
It is counter intuitive that the day of week plays the role that it does.
Best Answer
Hint:
If the couple have less than $2$ girls then they must have stopped after $2$ boys.
Possible scenarios: $BB$, $BGB$ and $GBB$.
Having at least $2$ girls is the complement of this.