1. Posterior p.f.

Suppose that the proportion Theta of defective items in a large manufactured lot is known to be either .1 or .2 and the prior p.f. of Theta is as follows:
E(.1)=.7 and E(.2)=.3
^Those E's are a greek letter in the text, I'm guessing representing a prior distribution.
Suppose also that when 8 items are selected at random from the lot, it is found that exactly two of them are defective. Determine the posterior p.f. of Theta.

Any help?? This is our first day doing this so I am very lost...

2. First, let X denote the number of defectives in our sample of 8. Then, X|theta ~ Bin(8, theta), clearly; using Bayes rule, we get
$\displaystyle
\xi(\theta|x) = \frac{\xi(\theta)f(x|\theta)}{f(x)}
$

With this, we now solve the problem in two steps. First, we need f(2), which we get via
$\displaystyle
f(2) = \sum_{\theta \in \{.1, .2\}} \xi(\theta)f(2|\theta).
$

Once you have this, just calculate xi(.1|2) and xi(.2|2), and you are done.