Plants and probabilities I am trying to work through some probability and statistics questions. We recently covered joint probability distributions but it is still quite fuzzy to me. I am not sure if this question (at least all of it) uses JPD's but here goes.
There are three varieties of a particular type of plant: variety A, variety B and a hybrid variety. 50% of the plants are of variety A, 20% of variety B and the remainder are hybrids.
(a) If a random sample of 10 plants is collected, nd the probability that the sample
contains:
(i) Exactly three plants of variety A;
(ii) Exactly two plants of variety B;
(iii) Exactly three plants of variety A and two plants of variety B.
In the sample, are the numbers of plants of each variety independent of each
other? Justify your answer clearly.
(b) Suppose now that the sample of 10 plants contains exactly four plants of variety
A. In this case, what is the distribution of the number of variety B plants in
the sample?

For part a), I think that for i and ii, it is enough to use Binomial but  I feel like there is a catch and that I will have to change the parameters for the 2 or 3 picks (namely, taking away one of the plants from the sample size and changing the probability accordingly). In iii I have to use Bayes from what I understand
For part b), I am not quite sure where to begin.
 A: For ($i$) the context seems to imply that these plants are coming from an infinite source. i.e. picking a variety A plant won't make you less likely to in the future. So we just have a standard binomial:
$${10 \choose 3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^7\approx0.1172$$
Similarly for ($ii$) we have
$${10 \choose 2} \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^8\approx0.302$$
For ($iii$) we have a multinomial distribution:
Let $X,Y,Z$ be random variables denoting the number of $A,B,$ and $C$ varieties. 
Then 
$$\begin{align*}
P(X=3, Y=2, Z=5)
&= \frac{10!}{3!\cdot2!\cdot5!}\left(\frac{1}{2}\right)^3 \left(\frac{1}{5}\right)^2 \left(\frac{3}{10}\right)^5\\\\
&\approx0.0306
\end{align*}$$
For ($iv$) no these events are not independent. For two events to be independent we need
$$P(A\cap B)=P(A)\cdot P(B)$$
but $$0.1172\cdot0.302\neq0.0306$$
($b$)
We are given that four of the plants are $A$. Now we know that the probability of a $B$ variety goes to $\frac{0.2}{0.2+0.3}=0.4$. Using the standard binomial as before and letting $X$ denote the number of variety $B$ plants, we get a pmf of
$$ p_{X}(x)=  
\begin{cases} 
0.046656 & x =0 \\
0.186624 & x=1 \\
0.311040 & x=2 \\
0.276480 & x=3 \\
0.138240 & x =4\\
0.036864 &  x=5\\
0.004096 & x=6\\
0 & otherwise
\end{cases} $$  
These calculations were made in R using the following command:
dbinom(0:6,6,.4) 
0.046656 0.186624 0.311040 0.276480 0.138240 0.036864 0.004096

A: In (a-i): Let $X \sim \mathsf{Binom}(n=10,p=.5).$ You seek
$P(X = 3).$ Use the formula for the binomial PDF (or PMF).
Using R statistical software, I got $P(X=3) = 0.1172,$ to four places. 
dbinom(3, 10, .5)
## 0.1171875

Similarly, in (a-ii) you have $X \sim \mathsf{Binom}(n=10,p=.5).$
dbinom(2, 10, .2)
## 0.3019899

I see that @Remy (+1) just posted more detail than I had intended to provide. So I will stop here.
A: The binomial distribution is a good approximation if the number of plants is large. We have to assume that. For iii) you don´t have to use the Bayes theorem. You have the multinomial distribution, where $A=3, B=2$ and $C=5$
For b) you can apply the Bayes theorem:
$$P(B=x|A=4,B+C=6)=\frac{P(A=4 \cap B=x\cap C=10-4-x)}{P(A=4)}$$
$$=\frac{\binom{10}{4,x,6-x}\cdot 0.5^4\cdot 0.2^x\cdot 0.3^{6-x} }{\binom{10}{4}\cdot 0.5^4\cdot 0.5^6}$$
