Question : We are given a coin which lands on head with probability p such that $ P(p \le x) = x^5$. Given that the first 9 tosses of the coin lands on head, what is the probability that the 10th toss is a head?
I have been practicing several probability problems, but having read this one I have no idea how to start. I am wondering if there is a way to determine a range on the value of p, but I do not have any idea how to use the information on the previous tosses. I would appreciate someone giving me some hints/some resources I can use to learn about how to solve these types of question.
One way to handle such problems is to use the notion of posterior probability. We begin with a distribution for a parameter (in your case, the probability) and update it as we see more data (in your case, coin tosses). The initial distribution is called the prior and the updated distribution is called the posterior.
Let $P$ be the probability that a coin flip is heads (a random variable). Let $T_{i}$ be an indicator random variable that is equal to one if and only if the $i$-th toss is heads. Fix a positive integer $n$ and let $\boldsymbol{T}=(T_{1},\ldots,T_{n})$. By the definition of conditional probability, \begin{equation} \mathbb{P}(P\leq p\mid\boldsymbol{T}=\boldsymbol{t})=\frac{\mathbb{P}(\boldsymbol{T}=\boldsymbol{t},P\leq p)}{\mathbb{P}(\boldsymbol{T}=\boldsymbol{t})}.\label{eq:conditional_cdf} \tag{conditional CDF} \end{equation} Let $\boldsymbol{1}=(1,\ldots,1) \in \mathbb{R}^n$ be the vector of ones, corresponding to the event in which the first $n$ tosses were all heads. Note that $$ \mathbb{P}(\boldsymbol{T}=\boldsymbol{1})=\int_{0}^{1}x^{n}\times5x^{4}dx=\frac{5}{n+5} $$ and $$ \mathbb{P}(\boldsymbol{T}=\boldsymbol{1},P\leq p)=\int_{0}^{p}x^{n}\times5x^{4}dx=\frac{5p^{n+5}}{n+5}. $$ Substituting these into the \ref{eq:conditional_cdf}, we get $$ \mathbb{P}(P\leq p\mid\boldsymbol{T}=\boldsymbol{1})=p^{n+5}. $$ The conditional density is obtained by differentiating: $$ f_{P\mid\boldsymbol{T}=\boldsymbol{1}}(p)=\left(n+5\right)p^{n+4}. $$ It follows that the probability that the ($n+1$)-th toss is a head is $$ \int_{0}^{1}pf_{P\mid\boldsymbol{T}=\boldsymbol{1}}(p)dp=\frac{n+5}{n+6}. $$
In the special case of $n=9$, this evaluates to $\boxed{14/15}\approx0.9\overline{3}$.
We can also use this to evaluate the unconditioned case (in which we have not seen any coin flips) by taking $n=0$. In this case, the probability is $5/6\approx0.8\overline{3}$.