binomial expansion questions Are the following answers correct and if not, how can I correct them?

(a) Write the expansion of $(2h+t)^5$.

I know Pascal's triangle has coefficients 1, 5, 10, 10, 5, 1
for a fifth degree binomial. So I should have $$1(2h)^5(t)^0+5(2h)^4t^1+10(2h)^3t^2+10(2h)^2t^3+5(2h)^1t^4+1(2h)^0t^5.$$
Simplifying this should give me $$32h^5+80h^4t^1+80h^3t^2+40h^2t^3+10ht^4+t^5.$$

(b) Consider the previous result as a description of flipping 5
  unfair coins. Even though the coin is unfair, which two outcomes are equally likely?

I'm not sure because normally the outcome is equally likely if the coin is fair but here I'm going to assume it's the outcomes with the same coefficients? I don't know why though. Also I haven't learned any probability in class and this was from a past exam I got wrong. 

(c) Consider again $(2h+t)^{100}$. Imagine you added every coefficient
  in the expansion. What would be the total?

So it would be 
$$\sum_{n=0}^{100} 2^n*1^{100-n}.$$
 A: 
(b) Consider the previous result as a description of flipping 5 unfair coins. Even though the coin is unfair, which two outcomes are equally likely?

Your guess is correct in question (b)
In the expression $(2h+t)^5$, $h$ represents a head and $t$ represents a tail. $(2h+t)^5$ is actually a description of five identical unfair coins each has $\frac{2}{3}$ chance getting a head and $\frac{1}{3}$ getting a tail. In the expanded form: $$32h^5 + 80h^4t^1 + 80h^3t^2 + 40h^2t^3 + 10ht^4 + t^5$$
We can see that the probabaility of getting 4 heads and 1 tail or getting 3 heads and 2 tails are both $\frac{80}{32+80+80+40+10+1} = \frac{80}{243}$.

(c) Consider again $(2h+t)^{100}$. Imagine you added every coefficient in the expansion. What would be the total?

It is often useful to simplify the case and find the pattern first. From the (b) part, we can see that the coefficient of $32h^5 + 80h^4t^1 + 80h^3t^2 + 40h^2t^3 + 10ht^4 + t^5$ add up to $243$, which is $3^5$. By following this pattern, we can find out the answer of (c) is $3^{100}$.
You may want to know why the pattern is like this. Here is the justification. Let's consider$(2h+t)^5$ again. When we plug $h=1$ and $t=1$, the equation becomes $$32(1)^5 + 80(1)^4(1)^1 + 80(1)^3(1)^2 + 40(1)^2(1)^3 + 10(1)(1)^4 + (1)^5 = 243$$
It is now clear that we can find the sum of coefficient just by plugging 1 in all the variables. Therefore the answer will be $$(2(1)+1)^{100} = 3^{100}$$
