An exercise in bayesian statistics Can you give me some suggestion on how to solve this?
Let $X_1, ... X_{20} |\theta \stackrel{iid}{\sim} Bern(\theta)$ and suppose $\theta \stackrel{}{\sim} Beta(2,2)$, namely:   
$P[X_i = x | \theta] = \theta^x(1-\theta)^{1-x} \mathbb{1}_{0,1} (x)$ , and  $g(\theta) = 6\theta(1-\theta)\mathbb{1}_{[0,1]}(\theta)$  
(a) identify the posterior distribution of $\theta$, given $X_1=x_1, ... X_{20} = x_{20} $  
What I did:
$f_{\theta | x}(\theta | x) = \frac{6\theta(1-\theta)}{\theta^{\sum_{i=1}^{n}x_{i}}(1 - {\theta)^{n-\sum_{i=1}^{n}x_{i}}} 6\theta(1-\theta)}$
Is this wrong? What is the range?
(b) determine the predictive distribution of $X_{21}$, given the observed sample, namely $P[X_{21} = 1 | x_1,...x_{20}]$  
What I did:
$f_{x}(x) = \theta^{\sum_{i=1}^{n}x_{i}}(1 - {\theta)^{n-\sum_{i=1}^{n}x_{i}}6\theta(1-\theta)}$
(c) If the observed sample is such that $s_{20} = \sum_{i=1}^{20} x_i = 10$, determine $P[\theta < 0.5 |  x_1,...x_{20}]$
What I did:
Should I compute the integral or the sum of $f_{x}(x)$? Or something totally different?

 A: If the individual tries $X_i$ are Bernoulli distributed, then their sum $k:=\sum_{i=1}^{20} X_i$ is binomially distributed and the probability density function is
$$
f(k|\theta,n) = \binom{n}{k}\theta^k(1-\theta)^{n-k}
$$
where the number of Bernoulli tries $n=20$ in your case.
What we are looking for is
$$
f(\theta|k,n) = \frac{f(k|\theta,n)f(\theta)}{f(k|n)}
$$
We also know $f(\theta)$ which is given by the beta distribution. Lets look at the nominator to decide what $f(k|n)$ needs to be so that $f(k|\theta,n)$ integrates to one:
$$\begin{align}
f(k|\theta,n)f(\theta) 
&= \binom{n}{k}\theta^k(1-\theta)^{n-k} \cdot \frac{1}{B(a,b)}\theta^{a-1}(1-\theta)^{b-1}\\
&=\binom{n}{k}\frac{1}{B(a,b)}
\theta^{k+a-1} (1-\theta)^{n-k + b -1}
\end{align}$$
This is almost the Beta distribution $Beta(a+k,b+n-k)$ and only differs by a constant factor. Since the pdf of the Beta distribution integrates to $1$ we know this constant factor musst be $f(k|n)$. Hence, to answer (a), the posterior distribution of $\theta$ is
$$
Beta\left(2+\sum_{i=1}^{20}X_i, 2 + 20 - \sum_{i=1}^{20}X_i\right).
$$
In order to answer (b) I assume by "predictive distribution" you mean the probability that $X_{21}=1.$ In general we need to integrate the statistic we are looking for time its pdf given by $Beta$:
$$
\mathbb E[\theta] = \int_0^1\theta \cdot f(\theta) \mathrm d \theta.
$$
This is just the expected value of the distribution and known for the beta distribution:
$$
\mathbb E[\theta] = \frac{2+\sum_{i=1}^{20}X_i}{2+2+20}
$$
Now (c) is just the value of the cumulative beta distribution at $0.5$ with $\alpha = 12$ and $\beta = 12$:
$$
\int_0^{0.5} f_{Beta}(\theta|12,12) \mathrm d \theta = 0.5
$$
