Basic Statistics, confused. Events There are two questions that I am having problems with.
The first one is  this one: 
             Event A: an even number turns up

             Event B: a 5 or 6 turns up

A die is rolled once: The probability of getting a 1 is 1/21,
 a 2 is 2/21, a 3 is 3/21, a 4 is 4/21,a 5 is 5/21, and a 6 is 6/21. 
They ask you to find P(A), P(B), P(AUB), P(AnB), P(AlB), P(BlA), and if the events are independent. I have an idea of how to do it and my results were: 
P(A): 4/7 ---- P(b): 11/21 ----P(AnB): 6/21----- P(AUB) 17/21 -----P(AlB): 6/11 ------ P(BlA): 1/2 and the events to not be independent because (AnB) =/= P(A) x P(B)
I am not exactly sure if I'm right. 
The second one is this one:
An unfair coin (Probability of 0.6 for heads) is flipped twice. Given that on at least one flip a head occurred, what is the probability that a head occurred on both flips. 
So in my mind, this question is really simple, but the wording makes me believe that I am overlooking something and that it is a bit more complicated. am I supposed to do like an x chooses n / binomial formula thing. 
Thank you in advance to anyone that can help me.
 A: Confirmations for die:
$$P(A) = P\{2,4,6\} = (2+4+6)/21 = 12/21 = 4/7.$$
$$P(A \cap B) = P\{6\} = 6/21 = 2/7,$$
$$P(B|A) = P(A \cap B)/P(A) = 2/4 = 1/2.$$
Binomial computations for coin:
Random variable $X \sim Binom(2, .6)$ is the number of Heads in two
tosses. Here there are $n = 2$ 'trials' with $P(\text{Success}) = P(\text{Heads}) = p = 0.6.$ Denote $q = 1-p.$ @justi is correct that using a binomial
random variable could get cumbersome. But this is a good place to
get started using the binomial distribution in a simple application. You want
$$P(X = 2 | X \ge 1) = \frac{P(X = 2,\,X\ge 1)}{P(X \ge 1)}
= \frac{P(X=2)}{P(X\ge 1)} = \frac{p^2}{2pq + p^2}.$$
The comma in the numerator of the first fraction is commonly-used notation for
$\cap$ between two expressions involving random variables. How do you justify the second equal sign? (If $A \subset B$, what is $A \cap B?$)
In R statistical software, where dbinom(k, 2, .6) computes 
$P(X = k) = {2 \choose k}p^k q^{n-k},$ for $k = 0, 1, 2.$
num = dbinom(2, 2, .6)
den = dbinom(1, 2, .6) + dbinom(2, 2, .6)
num/den
## 0.4285714

Note: It seems you are at the beginning of a very nice probability course.
You may find it useful to install R on your Windows, Mac, or Linux
computer, easy and free of charge from r-project.org. There is a lot to R
and way too much if you try to learn everything. But just getting
familiar with a dozen probability functions such as dbinom may
save you some time doing tedious computations.
