An interval for a proportion

You will write the 95% interval for a proportion, with a check of its conditions.

The idea. The estimate is $\hat p = \dfrac{\text{count}}{n}$. The margin is $1.96\sqrt{\dfrac{\hat p(1 - \hat p)}{n}}$. The formula is trusted only with at least $10$ successes and $10$ failures.

Your tasks.

1. Finish margin(count, n). Return $1.96$ times the standard error.
2. Finish interval(count, n). Return a pair (low, high).
3. Finish conditions_ok(count, n). Return True when there are at least $10$ successes and at least $10$ failures.