Minimizing the Sum of Binomial Random Variables 
5. A mountain climber has been lost on a mountain on either slope A or slope B. The head of the rescue mission believes that the mountain climber is on A with probability 0.6. Suppose that she has 12 rescue parties which are all equally competent and will locate the person with probability 0.3 if they are on the correct slope. Suppose further that they work independently of each other.
(a) If the mission leader sends eight parties to slope A and four to slope B, what is the probability that they rescue the lost mountain climber?
(b) How many rescue parties should she send to slope A in order to maximize the probability that the mountain climber is rescued?

I interpreted part (a) of this question as the weighted sum of two binomial random variables for success at finding the climber:
X~Bin(8, 0.3) and Y~Bin(4, 0.3).
By finding the probability that at least one group found him, I got the result:
0.6[1 - P(X=0)] + 0.4[1 - P(Y=0)] = 0.6(0.94235) + 0.4(0.7599) = 0.86937
However, I had a couple of problems with part (b):
1) Can you still use the sum X + Y = Z~Bin(12,0.3) since the probabilities of X and Y occuring are not the same?
2) If not, the simplest solution I can think of is to minimize the sum 
0.6*P(X = 0) + 0.4*P(Y = 0)
with the constraint that for X~Bin(k, 0.3) and Y~Bin(m, 0.3), k + m = 12. Is this a correct solution? And if so how do I approach it? I have found plenty of examples of maximizing a binomial random variable, but none about minimizing one. Thanks for the help!
 A: Here's how I would think about this problem: we can think of the probability of finding the climber as the probability of finding the climber given she went to mount A (times the probability she went there), plus the probability of finding the climber given she went to mount B (times the probability she went there). That is, 
\begin{align*}
P(\text{find})&=P(\text{find}\ A)P(A)+P(\text{find}|B)P(B)= P(\text{find}| A)0.6+P(\text{find}|B)0.4\\
&= (1-P(\text{no find}|A))0.6+(1-P(\text{no find}|B))0.4\\
&=(1-0.7^a)0.6+(1-0.7^b)0.4\\
&= (1-0.7^x)0.6+(1-0.7^{(12-x)}0.4),
\end{align*}
where $a$ denotes the number of search parties deployed to mount $A$, and $b$ to $B$, so $a+b = 12$; the last equality follows from $x=a$, so $b=12-x$. Note the third line follows from the fact that $P(\text{no find}|A)= P(\text{1 doesn't find}|A)*P(\text{2 doesn't find}|A)*\cdots * P(\text{a doesn't find}|A)=(0.7)^a$.
To find the find the number of parties that should be sent to mount $A$, maximize the expression above in $x$, which will tell you that they should send $x=7$ parties.
