Probability that last child is a boy Johnny has 4 children. It is known that he has more daughters
than sons. Find the probability that the last child is a boy.
I let A be the event that the last child is a boy, P(A) = $\frac{1}{2}$.
and B be the event that he as more daughters than sons. But im not sure how to calculate P(B) and what are the subsequent steps to take after.
Appreciate any help.
Thanks
 A: There are $2^4=16$ possible permutations of children, e.g. MMMM, or MFFM, or FFFM, or FFFF (here order is important, hopefully it's clear that MFFF means the first  child is male, the second is female, and so on) and each is equally likely. Now it just becomes conditional probability.
Let $A$ be the event that the last child is male. Let $B$ be the event that there are more female children than male children. The probability we are looking for is $P(A|B)$, the probability of $A$ given $B$. This is given by
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
where $A\cap B$ is the event where both $A$ and $B$ occur. Let's look at the terms individually:


*

*$A\cap B$ occurs precisely when Johnny has more daughters than sons, and the last child is a son. It's not hard to see there's only one possibility here: FFFM. Hence $P(A\cap B)=\frac1{16}$.

*$B$ occurs either when Johnny has no sons (one option, FFFF) or one son (four options, MFFF, FMFF, FFMF and FFFM). There are five options in total, so $P(B)=\frac5{16}$.


Putting this all together we get
$$P(A|B)=\frac{\frac1{16}}{\frac5{16}}=\frac15$$
so the probability that Johnny's last child was a son is one in five.
A: If he has more daughters than sons, Below are the 5 possible cases:   
D D D D --> All Daughters
S D D D --> 3 Daughters
D S D D
D D S D
D D D S    
So probability of having last child as son is = 1/5.
A: The number of girls in the family would have a binomial distribution, so the prior probability that there are 3 or 4 girls in the family would be:
$$\begin{align}
\mathsf P(B) & = {4\choose 3}(\tfrac 1 2)^3(\tfrac 1 2)+{4\choose 4}(\tfrac 1 2)^4
\\ & = \frac 5{16}
\end{align}$$
Now for the probability that the last child in the family is a boy and that there are more girls than boys in the family is equal to: the prior probability that the first three children are girls and the last is a boy:
$$\begin{align}
\mathsf P(A\cap B) & = \frac{1}{16}
\end{align}$$
Thus the posterior probability, that the last child is a boy given that their are more girls in the family than boys is:
$$\begin{align}
\mathsf P(A\mid B) & = \frac{\mathsf P(A\cap B)}{\mathsf P(B)}
\\ & = {\frac 1 {16}}\bigg/\frac 5 {16}
\\ & = \dfrac 1 5
\end{align}$$
A: The probability that any birth is a boy or a girl is NOT 1:1 as many people believe; in actuality 105 boys are born for every 100 girls. This ratio of 1.05 is known as the "secondary sex ratio." Given these real world statistics, one must give 1.05 weight to the four scenarios that include one boy and 1.00 weight to the all-girl possibility.
Therefore, the answer would be 1.05/5.20, or 21/104.
A: Picking up Robert Israel's suggestion: If a couple keeps having children until they have at least a boy and a girl, and ignoring twins, ... it's difficult. 
If they had children until they reached their goal, the probability is 1/16 each that they stop at 3 boys and 1 girl, at 1 boy and 3 girls, 3/4 that they stopped with 2 or 3 children, 1/8 that they had five or more. 
But they may not be finished yet. They may have four sons or four daughters and be waiting for another one. After being married for some time, there is a probability p that they have enough time for four children, and a probability q < p that they have enough time for five children. Both p and q grow with time, but don't reach 1. 
The probabilities if there are four children: p/16 for 1B + 3G, p/16 for 3B + 1G, (p - q)/16 each for 4B or 4G. Since there are more girls, p/16 for 1B + 3G and the last is a boy, (p-q)/16 for four girls. With conditional probabilities, the probability that the last is a boy is p / (2p - q). 
It would depend on the length of the relationship. If they are together for four years, it would be quite unlikely that they could have five children, (q would be small compared to p) and the probability of GGGB would be only slightly larger than GGGG, so the probability would be only a bit higher than 0.5. If they are together for many years, having GGGG and no fifth child is unlikely, so the chance that the last one is a boy is closer to 1. 
That also means that statistics for currently living parents and historical data would show different numbers. 
Obviously for different parental strategies the result would be different. 
A: They are mutually exclusive events. How many son/daughters the couple already has is irrelevant. Hence, the probability of the last child (or rather any child) being boy is 1/2 (some would say 1/3 too).
