okay.I think I got it this time.
The probability distribution actually makes sense since
P(N=0)+P(N=1)+...=(1-p)(1+p+p^2+p^3+...)=1
N:number of flowers
R:number of apples
By Bayesian rule,P(N=n|R=r) = P(R=r|N=n)*P(N=n) / P(R=r) (1)
P[N=n]= (1-p)p^n (2)
P(R=r|N=n)=C(n,r)a^r (1-a)^(n-r) (3) -there is no unknowns in this equation
Now we only need to solve P(R=r).
Notice that P(R=r) = P(R=r|N=n)*P(N=n) + P(R=r|Nn)*P(Nn) (4) - first term is also known by (2)&(3)
And P(Nn)=1-P(N=n) = 1 - (1-p)p^n (5)
Now we only need to solve P(R=r|Nn).Notice N>=r and n>=r.
P(R=r|Nn) = P(R=r|N=r)+P(R=r|N=r+1)+...+P(R=r|N=n-1)+P(R=r|N=n+1)+ ...
= -P(R=r|N=n)+P(R=r|N=r)+P(R=r|N=r+1)+...
=-C(n,r)a^r (1-a)^(n-r) + [P(R=r|N=r)+P(R=r|N=r+1)+...] (6)
Now we only need to solve P(R=r|N=r)+P(R=r|N=r+1)+...
P(R=r|N=r)+P(R=r|N=r+1)+...=a^r+C(r+1,r)a^r(1-a)+.C(r+2,r)a^r(1-a)^2
=a^r[1+C(r+1,r)(1-a)+C(r+2,r)(1-a)^2+...]
=a^r * [1+(1-a)]^r (7) - using Taylor's expansion to get this
Finally,plug(2)-(7) to (1) would solve the problem.
Hope this helps!