Four conditions, then the formula

A random variable is binomial when all four of these hold:

1. Fixed number of trials. $n$ is decided before any trial runs, not by what happens during them.
2. Two outcomes per trial. Each trial is classified as success or failure.
3. Constant probability. $p$ is the same on every trial.
4. Independent trials. One trial's result does not change another's probability.

Each condition rules out a real situation. "Shoot until you make one" has no fixed $n$, so it is not binomial. Rolling a die and recording the face has six outcomes, though "rolling a 6 or not" over 10 rolls is binomial. Drawing 5 cards from a 52-card deck without replacement changes $p$ from draw to draw, so it fails the last two conditions, and only the 10% condition rescues the approximation when the population is large relative to the sample.

State $n$ and $p$ explicitly whenever the model is invoked. A calculation that never named them usually skipped the check that would have shown the model does not apply.

For a binomial variable $X$ with $n$ trials and success probability $p$,

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \qquad \binom{n}{k} = \frac{n!}{k!(n-k)!}.$$

The two powers give the probability of one specific arrangement with $k$ successes; the coefficient counts how many arrangements produce exactly $k$. Both parts are required, and dropping the coefficient is the most frequent mechanical error in the topic.

A player makes 70% of her free throws and takes 5 shots. The probability she makes exactly 3 is

$$P(X = 3) = \binom{5}{3}(0.7)^3(0.3)^2 = 10(0.343)(0.09) = 0.3087.$$

Without the 10, the answer would be 0.03087, low by a factor of ten, because it would count only the one specific sequence make-make-make-miss-miss instead of all ten orderings.

For a binomial random variable,

$$\mu_X = np, \qquad \sigma_X = \sqrt{np(1-p)}.$$

For the 5 free throws at $p = 0.7$: $\mu = 3.5$ makes and $\sigma = \sqrt{5(0.7)(0.3)} = \sqrt{1.05} \approx 1.02$ makes. As always, the mean need not be attainable: 3.5 makes is a long-run average over many sets of five shots.

These formulas hold only when the binomial conditions hold. Applying $np$ to a count from a setting with a changing $p$, or with a random number of trials, produces a number with no meaning behind it, which is the reason the conditions come first in the write-up as well as in the reasoning.

A binomial question rarely asks for a single value. It asks for a range, and the translation is where answers go wrong:

1. At least 4 means $X \ge 4$, so $P(4) + P(5)$. For the free throws that is $0.36015 + 0.16807 = 0.52822$.
2. More than 4 means $X > 4$, so $P(5)$ alone.
3. At most 2 means $X \le 2$, so $P(0) + P(1) + P(2) = 0.16308$.
4. At least one is the complement of none: $P(X \ge 1) = 1 - P(X = 0)$.

The complement is a shortcut, not an alternative rule, and it must be taken against the correct event. The complement of "at least 3" is "at most 2", not "at most 3": every value from 0 to $n$ belongs to exactly one side. Writing the list of values before adding, and confirming the two sides together cover $0$ through $n$, catches nearly all of these.