Simulate n independent Bernoulli(p_true) trials and return (estimate, lower, upper): the observed proportion and a 95% confidence interval for it.

$$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \qquad CI = \hat{p} \pm 1.96\,SE$$

An estimate without an interval is close to useless, because it gives no way to tell a converged answer from a noisy one. Note the interval narrows with $\sqrt{n}$, so a tenfold improvement costs a hundred times the simulations.

Seed np.random.seed(seed) inside solve, immediately before drawing.

Input

p_true = 0.3
n = 10000
seed = 0

Output

(0.3055, 0.29647187000758185, 0.31452812999241814)