Consider a sequence of independent bernoulli random variables $X_1 X_2 ... X_n$ with parameter $\theta$ and $0<\theta<1$,where $P(X_i=1)=1-P(X_i=0)=\theta$. Assume the prior of $\theta$ follows Beta(3,2). Given that the first 9 observations are such that $\sum_{X_i = 1}^9 X_i = 7 $. what would be the (Bayesian) probability that $X_{10} =1$?
So I find out the posterior probability...
$P( \theta |X) \propto \theta^9(1-\theta)^3 $
And question ask me that $P(X_{10} = 1| H=7, T=2)$
But I don't have any idea for solving this question...
Hints:
Your posterior distribution for $\theta$ with density proportional to $\theta^9(1-\theta)^3$ is another Beta distribution. You should try to find its parameters
The posterior probability that the next value will be $1$ is then equal to the posterior expected value of $\theta$
One approach (not the quickest if you are familiar with Beta distributions) might be to find $$\dfrac{\int\limits_0^1 \theta \cdot \theta^9(1-\theta)^3\, d \theta}{\int\limits_0^1 \,\,\,\,\,\,\, \theta^9(1-\theta)^3\, d \theta}$$