Source: (Harvard Statistics 110: see #45, p. 28 of pdf).
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}
Question:
- 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. $$P(Y=k|\underbrace{X=15, B}_{\text{intersection removed}})P(B|X=15)+P(Y=k|\underbrace{X=15, B^c}_{\text{the same here}})P(B^c|X=15)$$ $$=P(Y=k|B)\ \ \ \ \ \ P(B|X=15)\ \ \ \ \ + \ \ \ \ \ \ \ \ P(Y=k|B^c) \ \ \ \ \ P(B^c|X=15)$$