The Binomial Distribution
▶︎ Watch it animatedinteractive step-through · ~3 min · optionalA binomial variable counts successes in a fixed number $n$ of independent trials, each with two outcomes and the same success probability $p$; all four conditions have to hold before the model is used. Then $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, where the powers give one arrangement's probability and the combination counts the arrangements, and $\mu = np$ with $\sigma = \sqrt{np(1-p)}$. For 5 free throws at $p = 0.7$: $P(X = 3) = 10(0.343)(0.09) = 0.3087$, $\mu = 3.5$, $\sigma \approx 1.02$.
Two failures dominate. The setting is never checked, so the model is applied where the number of trials is itself random (flip until heads), where $p$ changes trial to trial (drawing from a small group without replacement), or where the outcome is not two-valued. And boundaries are mistranslated: at least 4 summed as $P(4)$ alone, at most 2 summed from the wrong end, the complement of at least 3 taken as at most 3, or the combination coefficient dropped so that $P(X = 3)$ comes out ten times too small.
The work
3 ways in · any order
Lesson
The Binomial Distribution
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Puts the four binomial conditions ahead of the formula, then works the probability formula with its combination coefficient, the mean and standard deviation, and the translation of at least, more than, and at most into the correct sum.
Diagnostic
10-item topic check
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Ten items on the binomial model: settings that never qualified, the counting coefficient dropped, at least summed as exactly, and complements taken against the wrong event. Take it cold to find your habit, or after the lesson to check it is gone.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.