General hints regarding the use of the binomial distribution in conditional probabilty problems This is from an online course textbook (Harvard Statistics 110: see #45, p. 28 of  pdf). It's not a homework question, as the answer is provided in the text. I just really want to understand why the solution is that way.

A new treatment for a disease is being tested, to see whether it is better than the standard treatment. The existing treatment is effective on $50\%$ of patients. It is believed initially that there is a $2/3$ chance that the new treatment is effective on $60\%$ of patients, and a $1/3$ chance that the new treatment is effective on $50\%$ of patients. In a pilot study, the new treatment is given to $20$ random patients, and is effective for $15$ of them.  
(a) Given this information, what is the probability that the new treatment is better than the standard treatment?  
(b) A second study is done later, giving the new treatment to $20$ new random patients. Given the results of the first study, what is the PMF for how many of the new patients the new treatment is effective on? (Letting $p$ be the answer to (a), your answer can be left in terms of $p$.)

The answer to (b):
   Let $Y$ be how many of the new patients the new treatment is effective for and $p=P(B|X=15)$ be the answer from (a).  Then for $k\in \{0,1,\ldots,20\}$,
   \begin{align}
P(Y=k|X=15) &= P(Y=k|X=15,B)P(B|X=15)  \\
            &\quad + P(Y=k|X=15,B^c)P(B^c|X=15)  \\[5pt]
            &= P(Y=k|B)P(B|X=15) + P(Y=k|B^c)P(B^c|X=15)  \\[5pt]
            &= {20\choose k} (0.6)^k(0.4)^{20-k}p + {20\choose k}(0.5)^{20}(1-p).
\end{align}

Questions: 


*

*Before I looked at the answer, I didn't think that the answer to (a) would be found using a binomial distribution. But after I read the answer I was convinced by it. So, is there a kind of "general hint" when to use a binomial distribution?

*I don't understand why the answer to (b) is that way. Specifically, why is $$P(Y=k|X=15, B)=P(Y=k|B)$$ and the analogous one conditioning on $B^c$?
This equality is implied from the second equality of (b)'s answer.
$$P(Y=k|\underbrace{X=15, B}_{\text{intersection removed}})P(B|X=15)+P(Y=k|\underbrace{X=15, B^c}_{\text{the same here}})P(B^c|X=15)$$
$$=P(Y=k|B)\ \ \ \ \ \  P(B|X=15)\ \ \ \ \ + \ \ \ \ \ \ \ \ P(Y=k|B^c) \ \ \ \ \  P(B^c|X=15)$$
 A: 
Specifically, why is $$P(Y=k|X=15, B)=P(Y=k|B)$$



*

*The expression P(Y=k|X=15,B) = P(Y=k|B) is allowed because
$$P(Y=k|X=i,B) = {{20}\choose {k}} 0.6^k0.4^{20-k} $$ 
independent from $X=i$, so it can be left out
This logic is indeed not generally true, ie $P(a|b,c)$ may be different from $P(a|c)$, and is just true for this specific case where the experiments are independent (new random patients) and it is assumed that the result have no correlation other than that they are both defined by the condition $B$ or $B^c$.
Comparable expression with a dice roll for any $k$ and $l$: $$P(\text{next roll} =k \, | \,\text{ last roll} = l, \text{ dice = fair })=\frac{1}{6}$$
when we assume the rolls are independent. For that dice roll you are allowed to write $P(k|l,fair) = P(k|fair) = 1/6$

Before I looked at the answer, I didn't think that the answer to (a) would be found using a binomial distribution. But after I read the answer I was convinced by it. So, is there a kind of "general hint" when to use a binomial distribution?



*

*Effectively you have a mixture distribution of two binomial distributions 
$$ p \underbrace{ {{20} \choose {k}} 0.6^k0.4^{20-k}}_{\text{Binom$(20,0.6)$}} + (1-p) \underbrace{ {{20} \choose {k}} 0.5^k0.5^{20-k}}_{\text{Binom$(20,0.5)$}} $$
being equivalent to a probability $p$ that the medicine works on 60% of the people and a probability $1-p$ that the medicine works on 50% of the people. 

*This logic may not seem so correct. How could one, based on question 'a', believe that you have the medicine working on only either 50% or 60%? That is a false dichotomy. 
However, note that this is created due to the weird prior believe that it
is either working on 50% or 60% (probabilities 2/3 vs 1/3) and
nothing else (so the posterior distribution will stick to this weird
believe and only change these ratios 2/3 and 1/3 into $p$ and $1-p$).
It is not a realistic question, what one would get in practice. But it illustrates a more complicated principle (see next point).

*Eventually (a better alternative to the question problem statement which may occur later in your course), one could use a beta-binomial distribution in which every probability has an assigned value (not just 2/3 60% and 1/3 50%, or $p$ 60% and $1-p$ 50%). Then the posterior distribution for $f$, the fraction of people on which the medicine works (which could be modeled as a beta distribution), is a spectrum instead of just two values. This spectrum of probabilities for different $f$ is then used to write out the beta-binomial distribution.

*note that the letter $p$ is often a parameter for the binomial distribution, the probability that the medicine works. Here it is used differently for the probabilities $p$ and $1-p$ that the medicine works with 60% or 50%. This might be confusing.
A: In this response I will limit myself to the general question of when to use the binomial.
The binomial arises naturally as the count of successes in a sequence of independent Bernoulli trials with constant success probability $p$.
It's useful to keep the conditions in mind (both for Bernoulli trials and the independent and constant $p$ of the sequence of them) and see whether that might be more-or-less plausible. [Often the model is a reasonable approximation when the conditions don't quite hold.]
For example if you have a population with some fraction having some attribute and you sample it at random with replacement, the count of the number having the attribute in your sample would be an obvious candidate for using a binomial. If you instead sample without replacement but the population is very very large compared to your sample (making the effect on $p$ at trial $i$ and the dependence negligible) then you could use the binomial as an approximation.
