Compute exact probabilities for the binomial distribution including $P(X = k)$, $P(X \le k)$, $P(X \ge k)$, and use complementary counting

What this dot point is asking

NESA wants you to compute binomial probabilities of every type: exact $P(X = k)$, ranges $P(X \le k)$ or $P(X \ge k)$, and at-least/at-most using complementary counting.

The answer

Exact probability

For $X \sim B(n, p)$,

Cumulative probability (sum of pmf values)

For HSC problems with small $n$, sum the pmf values up to $k$.

Complementary probability

For "at least $k$":

For small $k$ (like $k = 1$ or $k = 2$), this is much faster than summing $k$ to $n$.

Standard problem patterns

Exact number of successes: $P(X = k)$ directly from the pmf.

At least one success: $P(X \ge 1) = 1 - P(X = 0) = 1 - (1 - p)^n$.

No successes: $P(X = 0) = (1 - p)^n$.

All successes: $P(X = n) = p^n$.

Between two values: $P(j \le X \le k) = \sum_{i = j}^{k} P(X = i)$.

Choosing the easier sum

If asked $P(X \ge k)$ and $n - k + 1$ is small, sum directly. If $n - k + 1$ is large, use $1 - P(X \le k - 1)$ (complementary).

For "at most $k$", consider whether $k$ or $n - k$ is smaller. If $k$ is smaller, sum directly.

Complementary counting, stage by stage

The single most useful move in these questions is to count the complement when it is shorter. Take a five-question multiple-choice test with four options each, guessed at random, so $X \sim B(5, 0.25)$, and suppose the question asks for $P(X \ge 2)$, the probability of getting at least two right. The three panels show why the complement wins.

Stage 1, picture the whole distribution. The bars are the six probabilities $P(X = 0)$ through $P(X = 5)$ from the pmf. Getting "at least $2$" right means the bars from $k = 2$ rightward, which is four terms to add. That is doable but slow.

Stage 2, identify the shorter complement. The complement of "at least $2$" is "fewer than $2$", that is $X = 0$ or $X = 1$, the two leftmost bars (shown in accent). Only two terms: $P(X = 0) = (0.75)^5 \approx 0.2373$ and $P(X = 1) = \binom{5}{1}(0.25)(0.75)^4 \approx 0.3955$.

Stage 3, subtract from one. The answer is everything that is not the complement: $P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \approx 1 - 0.2373 - 0.3955 = 0.3672$. The accent bars now mark the region you actually wanted, found with two terms instead of four.

Why the binomial coefficient appears

The factor $\binom{n}{k}$ in the probability formula counts the number of distinct orderings in which exactly $k$ successes can fall among $n$ trials. Each specific ordering has probability $p^k (1 - p)^{n - k}$, and there are $\binom{n}{k}$ of them, so the total probability of exactly $k$ successes multiplies the two. This is also why binomial probabilities and the binomial theorem share the same coefficients: expanding $(p + q)^n$ produces exactly the terms $\binom{n}{k} p^k q^{n - k}$, and summing them over all $k$ gives $(p + q)^n = 1$, confirming the whole distribution has total probability one.

How exam questions ask about binomial probabilities

Half the difficulty in these questions is converting English into the right probability statement, then choosing the shorter calculation. The wordings, and the inequality each one means, are:

• "exactly $k$": $P(X = k)$ straight from the pmf.
• "at least $k$": $P(X \ge k)$. For small $k$ use the complement $1 - P(X \le k - 1)$.
• "at most $k$": $P(X \le k)$, a direct sum from $0$ to $k$.
• "more than $k$": $P(X \ge k + 1)$, one step further than "at least $k$".
• "fewer than $k$" / "less than $k$": $P(X \le k - 1)$, stopping one short of $k$.
• "between $j$ and $k$ (inclusive)": $P(j \le X \le k) = \sum_{i=j}^{k} P(X = i)$.
• "at least one": the most common complement of all, $P(X \ge 1) = 1 - (1 - p)^n$.

Misreading "more than" as "at least" shifts the boundary by one and is a frequent source of lost marks. Once the inequality is fixed, decide whether a direct sum or a complement is the shorter calculation, and state which you are using.