Two coins tossed $5$ times with different probabilites for heads on each coin. What's the probability of getting at least $3$ heads? Given two coins with $P_1(H) = \frac{1}{2}$ and $P_2(H) = \frac{1}{3}$, one is chosen at random and tossed $5$ times, what's the probability of tossing at least $3$ heads?  
If $A$ is the probability of getting at least three heads, then $P(A) = P(3) + P(4) + P(5)$ where $P(i)$ is the probability of getting $3, 4,$ and $5$ heads exactly respectively.  
I know see that $P(3) = C_3^5P(H)^3(1-P(H))^{5-3}$, but this depends on the coin being tossed.
I also know that $P(3) = P(3 | B_1)P(B_1) + P(3 | B_2)P(B_2)$ for some partition ${B_1, B_2}$.  But what would be an appropriate partition here?  
 A: Rather than determining the probability of getting $n$ heads as a function of the probability of heads, and then conditioning the probability of heads on which coin you chose, swap the order.
In other words, assume the probability of choosing coin $1$ or coin $2$ is $1/2$ either way.  Then, given you choose coin $1$, the probability of getting at least $3$ heads is trivially $1/2$ (do you see why?).  Given you choose coin $2$, the probability of getting at least $3$ heads is smaller, but (as you point out) can be obtained as
$$
P(\text{at least $3$ heads $\mid$ coin $2$})
    = \sum_{k=3}^5 \binom{5}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{5-k}
$$
This should be straightforward to compute.  Then we can compute
\begin{align}
P(\text{at least $3$ heads})
    & = P(\text{at least $3$ heads $\mid$ coin $1$}) P(\text{coin $1$}) \\
    & + \, P(\text{at least $3$ heads $\mid$ coin $2$}) P(\text{coin $2$})
\end{align}
A: Here are results from simulating this experiment a million times,
using R statistical software. The approximate probability for exactly three Heads from simulation
should be accurate to 2 or 3 places is shown. The exact probability of exactly three Heads is given
for comparison; it is based on partitioning on which coin is chosen. 
In R, dbinom is the binomial PDF and rbinom simulates
the binomial random variable described by parameters.
You can use the same method to find the probability of at least three Heads.
 m = 10^6;  x = numeric(m)         # x is nr of H's in 5 tosses
 for (i in 1:m) {
   p = sample(c(1/2, 1/3), 1)      # pick which coin: 50-50 chance
   x[i] = sum(rbinom(1, 5, p)) }   # simulate 5 tosses with that coin
 mean(x == 3)
 ## 0.238784                       # aprx P(X = 3) from simulation
 (dbinom(3, 5, 1/2) + dbinom(3, 5, 1/3))/2
 ## 0.2385545                      # exact P(X = 3)

This problem illustrates the 'Law of Total Probability' in a form often
used to prove 'Bayes' Theorem'.
The plot below shows the probabilities of the required mixture distribution (heavy purple bars), along with the binomial distributions from tossing
each of the two coins (thin red and blue lines adjacent).

A: 
The book I'm reading defines an event as a subset of the sample space, so if coin $1$ is an event then it must be a subset of the sample space. This subset must be all combinations of coin tosses given coin $1$, which is the same as the amount of coin tosses using coin $2$. What exactly is in the events (subsets of the sample space) coin $1$ and coin $2$ that makes them a partition is what I'm confused about. coin $1 = {x,y,z,...}$ and coin $2={w,r,t,...}$ such that coin $1 ∩ $ coin $2=∅$ and their union is the entire space of coin tosses. 

An appropriate sample space to describe this system is $\{\tfrac 1 2,\tfrac 1 3\}{\times}\{{\rm T,H}\}^5$, that is the Cartesian product of outcomes for selecting a coin (denoted by its bias), and tossing that coin five times.    Your random variable is the count of heads obtained.   Let us call it $X$.
$X=0$ denotes the event of $\rm \{(\tfrac 12, T,T,T,T,T),(\tfrac 13, T, T, T, T, T)\}$ .   That is the outcome of selecting the unbiased coin and then tossing five tails, or of selecting the biased coin and then tossing five tails.   And so forth.    The probability of this event is calculated as: $$\mathsf P(X=0)~=~ \tfrac 12(\tfrac 12)^5+\tfrac 12(\tfrac 23)^5$$
As you can see, we can partition the space as $\{\tfrac 12\}{\times}\{{\rm T,H}\}^5~\cup~\{\tfrac 13\}{\times}\{{\rm T,H}\}^5$, that is on the event of selecting which coin.   Let $Y$ be the bias of the coin.
$Y=\tfrac 12$ describes the event $\rm \{(\tfrac 12, T, T, T, T, T), (\tfrac 12, T, T, T, T, H), \ldots, (\tfrac 12, H, H, H, H, H)\}$, etcetera.   Thirty two outcomes of various probabilities.  $\mathsf P(Y=\tfrac 12)=\tfrac 12=\mathsf P(Y=\tfrac 13)$ because selecting the coin itself is made without bias.
That's the meat.   The crunch of it is that:
$$\begin{align}\mathsf P(X\geq 3) ~=~& \mathsf P(Y=\tfrac 12\cap X\geq 3)+\mathsf P(Y=\tfrac 13\cap X\geq 3)
\\[1ex] =~& \tfrac 12 \Big(\sum_{x=3}^5 \binom 5 x (\tfrac 12)^x(\tfrac 12)^{5-x}\Big)+\tfrac 12\Big(\sum_{x=3}^5 \binom 5 x (\tfrac 13)^x(\tfrac 23)^{5-x}\Big)
\\[1ex] =~& \tfrac 14+\tfrac 1 2 \Big(\binom 5 3\tfrac{2^2}{3^5}+ \binom 5 4\tfrac{2}{3^5}+\binom 5 5\tfrac 1{3^5} \Big)
\\[1ex] =~& \tfrac 14+\tfrac{40+10+1}{2\cdot 3^5}
\\[1ex] =~& \tfrac {115}{324}
\\[1ex] =~& 0.35\dot{\overline{493827160}}
\end{align}$$
