Large Counts Condition vs 10% Condition

Large Counts is a sample size check: you need roughly 10 successes and 10 failures before you can treat the sampling distribution of $\hat{p}$ (p-hat, the sample proportion) as normal; that is the textbook name for it, and the AP course framework calls the same check the normality condition. Which value of $p$ goes into the check depends on the procedure. A significance test has a hypothesized value, so you verify $np_0 \ge 10$ and $n(1-p_0) \ge 10$ using $p_0$ (p-naught); a confidence interval has no hypothesized value, so you verify $n\hat{p} \ge 10$ and $n(1-\hat{p}) \ge 10$, which is just counting the observed successes and failures. Testing $p_0 = 0.25$ with $n = 60$ passes, since $np_0 = 15$ and $n(1-p_0) = 45$.

Sampling without replacement makes each draw depend a little on the ones before it, because removing a person changes what is left in the population. The 10% condition, written $n \le 0.10N$ (n at most one tenth of N, the population size), keeps that dependence small enough that standard deviation formulas built for independent draws still hold. In a school of 1,200 students, $0.10(1200) = 120$, so a sample of 80 satisfies the condition while a sample of 200 does not. You only need this check when sampling without replacement.