Given an observed sample from a binomial distribution, how do I solve for the probability distribution of possible 'p' values?

Let's say I have an 'unfair' coin, for which I'm interested in estimating the 'heads' likelihood or 'p' value.

Knowing nothing about the coin, the distribution of probable 'p' values is a uniform distribution from 0 to 1.

How can I update this distribution of probable 'p' values after observing X heads out of Y trials?

So I'll add the details here. Let $$x_i$$, $$i=1,\ldots,Y$$ be the outcomes from coin tosses. Given: $$\sum_{i=1}^Y x_i = X$$, the number of heads. Let $$\theta = Pr(X_i = 1)$$, where we'll treat 'head' as 1. Therefore, the random variable $$X_i \sim Bernoulli(\theta)$$, with 'head' as the 'success'.
Given prior for $$\theta$$: $$f(\theta) = 1 I(\theta \in (0,1))$$, which is unif(0,1). Since $$\sum_{i=1}^Y x_i$$ is $$Binomial(Y,\theta)$$, the likelihood is $$Pr(\sum_{i=1}^Y x_i = X) = K\theta^X(1-\theta)^{Y-X}$$, where $$K$$ is $$Y$$ choose $$X$$.
Now, posterior of $$\theta$$: $$f(\theta\mid D) = \frac{f(D\mid\theta)f(\theta)}{\int_{0}^{1}f(D\mid\theta)f(\theta)d\theta}$$, where $$D$$ is $$\sum_{i=1}^Y x_i$$, the 'data'. Therefore, $$f(\theta\mid D) = \frac{K\theta^X(1-\theta)^{Y-X}}{K\int_{0}^{1}\theta^{X}(1-\theta)^{Y-X}d\theta} = \frac{\theta^X(1-\theta)^{Y-X}}{Beta(X+1, Y-X+1)}$$, which is the density function of the Beta distribution with parameters $$X+1$$ and $$Y-X+1$$. $$Beta(,)$$ is the Beta function.
I have gone through an example of updating the probability of $$p$$ after each observation, starting with a flat prior - with a graph of the posterior probability of $$p$$ after each update. Basically the same as the other 2 answers, but graphed it, and done sequentially after each new flip.