Geometric Distribution - Biased Coin Flip Whilst studying I stumbled on this problem, which I wish to check if my understanding is correct.
Imagine we have a biased coin with probability 'k' of getting a head when flipped.
Now A defines the number of times a biased coin is thrown until getting the 1st Head. On the other hand, we have B which represents the number of times a biased coin is thrown until getting the 1st tail.
Question: Show that $p(1,b) = k^{b-1}(1-k)$, where $p(a,b)$ is the joint distribution of a & b.

My Understanding:
From the problem I believe that we can say that A ~ Geo(1) & B ~ Geo(b).
Moreover, we can say that A & B are two dependent random variables as the number of flips in A affects the number of flips of B and vice versa. 
Now, to find $ p(1,b)$
$p(1,b) = P(A=1, B=b)$
$= P(B=b|A=1)P(A=1)$
{Using the Geometric distribution: $P(X=n)=(1-p)^{n-1}p$}
$= [(1-(1-k))^{(b-1)-1}(1-k)] [(1-k)^{1-1} (k)]$
$= [(1-(1-k))^{b-2}(1-k)][(1-k)^{0}(k)]$
$= k^{b-2}(1-k)(k)$
$= k^{b-1}(1-k)$
Can someone verify if my understanding is correct? Thanks :)

EDIT:
Just to make sure that I have understood the answer given correctly:
In case that this time we have $p(a,1)$
A~Geo(k)and B~Geo(1-k)
$p(a,1) = P(A=a, B=1)$
$P(B=1 | A=a)P(A=a)$
$=[(1-(1-k))^{(1-a)-1}(1-k)][(1-k)^{a-1}k]$
$=[k^{-a}(1-k)][(1-k)^{a-1}(k)]$
$k^{1-a}(1-k)^{a}$
 A: \begin{align}
p(1,b) &= P(A=1, B=b) \\
&= P(B=b|A=1)P(A=1) \\
&=P(B=b|A=1)k \\
&=\begin{cases}0 & , b=1 \\ k^{b-2}(1-k)k
&, b > 1\end{cases} \\
&=\begin{cases}0 & , b=1 \\ k^{b-1}(1-k)
&, b > 1\end{cases} \\
\end{align}


*

*Try to  compute $p(a,1)$ similarly as an exercise. 


Comment on your current answer to compute $p(a,1)$, well, the two problems are actually quite symmetrical, so the lack of symmetric in the answer that you obtained should warns us. 
Your mistake was actually when you deal with $P(B=1|A=a)$. If $a=1$, well, it is equal to zero since we can't get two outcome simultaneously. If $a>1$, well, the first toss is not head, hence the first toss must be tail. Hence in summary:
$$P(B=1|A=a)=\begin{cases}0 &, a=1 \\ 1 &, a > 1\end{cases}$$


*

*As another exercise, try to compute $p(1,b)=P(A=1|B=b)P(B=b)$ to  get the answer that I provided you earlier. 

*Remark: A sanity check, you want to make sure the joint pmf sums up to $1$. 
A: A variable X~Geo(p) where p is the probability of success for each trial. Since the probability of getting heads is 'k', 
A ~ Geo(k)
And since B is the complement of A, their probabilities must sum to one. So,
B~Geo((1-k)). The rest of your work seems correct.
