Help in evaluating conditional probabilities. I need some help in evaluating some probabilities. 
My probability course is slightly too theoretical and abstract for me- we have been shown identities and theorems relating these probabilities but never were we shown an example or given any insight into how to actually evaluate them, perhaps I lack the assumed intuition but going purely off the definitions given to me I have no way that I'm aware of of actually doing the following (very simple) question.
Ideally some insight rather than a full solution to the below would be great- perhaps an outline or the first few lines or even a link to a similar question and a solution to that that I can apply here. But I'm aware that's a lot to ask so anything at all is appreciated.

Parliament contains a proportion $p$ of Labour members, who are
  incapable of changing their minds about anything, and a proportion
  $1−p$ of Conservative members who change their minds completely at
  random (with probability $r$) between successive votes on the same
  issue. A randomly chosen member is noticed to have voted twice in
  succession in the same way. What is the probability that this member
  will vote in the same way next time?
  So presumably we start with (say they have voted for $A$ so far)

$\mathbb P($will vote $A$ again$)=\mathbb P($will vote $A\space|\space$voted for $A$ twice$)=\mathbb P($will vote $A\space\cap$ voted for $A$ twice)$/\mathbb P($voted for $A$ twice$)$ but I don't know how to now evaluate these probabilities. 
Thank you
 A: If you are finding the course too theoretical and abstract, perhaps you should try and understand the process rather than mechanically apply formulas.
I shall use assumed figures, I trust you can translate it into the formulas you have learned.
Labour MPs = 40, Conservative MPs = 60, who change their mind with $Pr = 0.4$ 
Number of Labour MPs voting A twice = 40,
Number of Conservative MPs voting twice $=60\times(1-0.4) = 36$
P(Labour |voted A twice) $=\dfrac{40}{76}$
P(Conservative | voted A twice) $=\dfrac{36}{76}$
P(will vote A again) $= \dfrac{40}{76}\cdot1 + \dfrac{36}{76}\cdot(1-0.4)$    
A: Well, this person is either one of those that are incapable of changing their minds, or it is one of the others that randomly change their minds.A: will vote for A
C: can change their mind
R: repeated their vote for A
So: 
$P(A|R) = P(A\land C|R) + P(A \land \neg C|R)$
where
$P(A\land C|R) = P(A|C,R)*P(C|R)$
$P(A\land \neg C|R) = P(A|\neg C,R)*P(\neg C|R)$
and where 
$P(A|C,R)=1-r$  (the mind-changer will vote for A again with probability $1-r$)
and
$P(A|\neg C,R)=1$ (the mind-non-changer will certainly vote for A again)
and where (Bayesian rule):
$P(C|R) = \frac{P(R|C)*P(C)}{P(R)}$
where
$P(R) = P(R \land C) + P(R \land \neg C)$
$P(R \land C) = P(R|C) * P(C)$
$P(R \land \neg C) = P(R|\neg C) * P(\neg C)$
and
$P(R|C) = 1-r$
$P(R|\neg C) = 1$
$P(C) = 1-p$
$P(\neg C) = p$
(I'll leave the Bayesian for $P(\neg C|R)$ to you)
