Define and apply the binomial distribution to model the number of successes in $n$ independent Bernoulli trials, including computing probabilities, expected value $np$ and variance $np(1-p)$

What this dot point is asking

VCAA wants you to recognise binomial situations, write the probability mass function, and use the standard formulas for the expected value and variance.

Bernoulli trial

A single experiment with two possible outcomes (success or failure) and probability $p$ of success. Examples: coin toss, single quality-control inspection, single penalty kick.

Binomial distribution

If $n$ independent Bernoulli trials are run with the same success probability $p$, then $X$ = number of successes is binomially distributed: $X \sim \text{Bin}(n, p)$.

Probability of exactly $k$ successes:

where $\binom{n}{k} = ^nC_k = \dfrac{n!}{k!(n - k)!}$.

Conditions

To apply the binomial:

• $n$ is fixed in advance.
• Each trial has two outcomes (success / failure).
• Probability of success is the same on every trial.
• Trials are independent.

Expected value and variance

The mean $np$ is intuitive: in $10$ tosses of a fair coin, expect $5$ heads. The variance peaks at $p = 0.5$ (most uncertain outcome).

Cumulative probabilities

$P(X \le k) = \sum_{j=0}^k \binom{n}{j} p^j (1 - p)^{n - j}$.

CAS calculators give cumulative probabilities directly via BinomialCDF(n, p, k).

Cumulative and "at least" probabilities

Most exam questions ask for ranges, not single values, so fluency with cumulative probabilities is essential. "At most $k$" is $P(X \le k)$, computed directly with the CAS binomCdf. "At least $k$" is best handled with the complement: $P(X \ge k) = 1 - P(X \le k - 1)$, which avoids summing a long tail. "Between $a$ and $b$ inclusive" is $P(X \le b) - P(X \le a - 1)$. The single most common boundary error is forgetting that $P(X \ge k) = 1 - P(X \le k - 1)$, not $1 - P(X \le k)$.

Finding $n$ or $p$ from a condition

VCAA also poses inverse problems: given a required probability, find the number of trials $n$ or the success probability $p$. The "at least one" structure is the classic case: $P(X \ge 1) = 1 - (1 - p)^n$, set against a target, then solved for $n$ using logarithms (since the unknown is an exponent). Solving for $p$ from a single-term equation such as $\binom{n}{k} p^k (1-p)^{n-k} = c$ is done with CAS solve, restricting to $0 < p < 1$ to discard non-physical roots.

In one sentence

The binomial distribution $X \sim \text{Bin}(n, p)$ models the number of successes in $n$ independent Bernoulli trials with success probability $p$, with $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, expected value $E[X] = np$ and variance $\text{Var}(X) = np(1 - p)$.

Examples in context

Example 1. Quality control on a production line. A factory's bottling line has a $5\%$ defect rate. In a random sample of $8$ bottles, let $X$ be the number defective, so $X \sim \text{Bin}(8, 0.05)$. The probability of exactly one defective is $P(X = 1) = \binom{8}{1}(0.05)^1(0.95)^7 = 8 \times 0.05 \times 0.6983 = 0.2793$. The expected number of defectives per sample is $E[X] = 8 \times 0.05 = 0.4$.

Example 2. Free-throw shooting. A basketballer makes $70\%$ of free throws. In $5$ attempts, $X \sim \text{Bin}(5, 0.7)$. The probability of making at least $4$ is $P(X \ge 4) = P(X = 4) + P(X = 5) = \binom{5}{4}(0.7)^4(0.3) + (0.7)^5 = 5(0.2401)(0.3) + 0.16807 = 0.36015 + 0.16807 = 0.5282$.

Try this

Q1. For $X \sim \text{Bin}(12, 0.25)$, find $E[X]$ and $\text{Var}(X)$. [2 marks]

• Cue. $E[X] = 12 \times 0.25 = 3$; $\text{Var}(X) = 12 \times 0.25 \times 0.75 = 2.25$.

Q2. A spinner lands on red with probability $0.4$. It is spun $6$ times. Find the probability of exactly $2$ reds. [3 marks]

• Cue. $\binom{6}{2}(0.4)^2(0.6)^4 = 15 \times 0.16 \times 0.1296 = 0.311$.

Q3. A test has $10$ true/false questions answered by guessing ($p = 0.5$). Find the probability of getting at most $1$ correct. [3 marks]

• Cue. $P(X = 0) + P(X = 1) = (0.5)^{10} + 10(0.5)^{10} = 11 \times \frac{1}{1024} = \frac{11}{1024} \approx 0.0107$.