probability, please help on bayes question I dont need exact answer, I just need help to judge wether my following method is correct or not.
Question:A physician has 5 patients. There are treatments A and B. Physician gives treament A to 3 randomly selected patients, and B to the other 2. Suppose that treament A produce a cure in any patient with a probability of 0.3, and B with 0.6. What is the probability that the two treatments will produce the same number of cures?
My attempt:
Let 
X= # of cure in treament A
Y=# of cure in treament B
P(X=Y)=P(X=0,Y=0)+P(X=1,Y=1)+P(X=2,Y=2)
=P(X=0)P(Y=0)+P(X=1)P(Y=1)+P(X=2)P(Y=2)
P(X=0)=(p)^3*(1-p)^2, but I need to figure out what is p. 
now comes to my question. For treament A, P(Cure|treament A)=0.3 as given in the problem. My question is should I use this number or P(Cure And treament A)=P(A)*P(Cure|treament A) to fill in the "p" in the above?

The setup is correct, we want 
$$\Pr(X=0)\Pr(Y=0)+\Pr(X=1)\Pr(Y=1)+\Pr(X=2)\Pr(Y=2).$$
We have $\Pr(X=0)=(0.7)^3$ and $\Pr(Y=0)=(0.4)^2$.
For $\Pr(X=1)$, the right expression is $\binom{3}{1}(0.3)(0.7)^2$. Similarly, $\Pr(Y=1)=\binom{2}{1}(0.6)(0.4)$.
Finally, $\Pr(X=2)=\binom{3}{2}(0.3)^2(0.7)$ and $\Pr(Y=2)=(0.6)^2$. 
Remark: There is nothing particularly Bayesian about the calculation. The random variables $X$ and $Y$ have binomial distribution, with parameters fully specified in the statement of the problem. 
