Bernoulli trials and the binomial distribution: VCE Math Methods Unit 3

What this dot point is asking

VCAA wants you to recognise when a situation is binomial, write down $X \sim \mathrm{Bi}(n, p)$, compute exact binomial probabilities by hand using $\binom{n}{x} p^x (1-p)^{n-x}$ for small $n$ on Paper 1, and use CAS on Paper 2 to compute cumulative probabilities. The mean $np$ and variance $np(1-p)$ are essential shortcuts.

Bernoulli trials

A Bernoulli trial is a single experiment with exactly two outcomes (commonly called "success" and "failure"), where the probability of success is some fixed $p$ (and probability of failure is $1 - p$).

A Bernoulli random variable takes the value 1 for success and 0 for failure, with probability distribution $P(Y = 1) = p$ and $P(Y = 0) = 1 - p$. Its mean is $p$ and its variance is $p(1 - p)$.

The binomial distribution

When $n$ Bernoulli trials are performed under the four binomial conditions, the total number of successes follows a binomial distribution.

The four conditions for a binomial

1. Fixed number of trials $n$.
2. Each trial has only two outcomes, success or failure.
3. The probability of success $p$ is constant across trials.
4. The trials are independent.

If any one of these fails, $X$ is not binomial. The most common failure is sampling without replacement (which breaks independence for finite populations).

Notation: $X \sim \mathrm{Bi}(n, p)$. The two parameters $n$ and $p$ fully determine the distribution.

The binomial probability formula

For $X \sim \mathrm{Bi}(n, p)$ and $x \in \{0, 1, 2, \dots, n\}$:

where $\binom{n}{x} = \frac{n!}{x! (n - x)!}$ counts the number of ways to choose which $x$ of the $n$ trials are successes.

The three factors have natural interpretations:

• $\binom{n}{x}$: number of distinct orderings of $x$ successes and $n - x$ failures.
• $p^x$: probability of $x$ specific successes occurring.
• $(1 - p)^{n - x}$: probability of the remaining $n - x$ trials being failures.

Worked example

A biased coin lands heads with probability $0.6$. It is tossed 5 times. Find the probability of exactly 3 heads.

$X \sim \mathrm{Bi}(5, 0.6)$.

$P(X = 3) = \binom{5}{3} (0.6)^3 (0.4)^2 = 10 \cdot 0.216 \cdot 0.16 = 0.3456$.

Mean and variance

For $X \sim \mathrm{Bi}(n, p)$:

These shortcut formulas come directly from the linearity rule applied to the sum of $n$ Bernoulli trials.

For the biased coin above: $E(X) = 5 \cdot 0.6 = 3$, $\mathrm{Var}(X) = 5 \cdot 0.6 \cdot 0.4 = 1.2$.

Cumulative binomial probabilities

For "at least", "at most" or "between" probabilities, you sum binomial terms.

• $P(X \leq k) = \sum_{x = 0}^{k} \binom{n}{x} p^x (1 - p)^{n - x}$
• $P(X \geq k) = 1 - P(X \leq k - 1)$
• $P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1)$

On Paper 1, $n$ is small enough to compute terms by hand. On Paper 2, use the CAS binomCdf(n, p, lower, upper) function (or its equivalent on your model).

Worked example

A fair die is rolled 10 times. Find the probability of at least 3 sixes.

$X \sim \mathrm{Bi}(10, 1/6)$ where $X$ is the number of sixes.

$P(X \geq 3) = 1 - P(X \leq 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]$.

$P(X = 0) = (5/6)^{10}$.

$P(X = 1) = 10 (1/6)(5/6)^9$.

$P(X = 2) = 45 (1/6)^2 (5/6)^8$.

Sum these (or use CAS) and subtract from 1. The result is approximately $0.2248$.

Worked example: solving for the parameter

A biased coin is tossed 10 times. The probability of getting exactly 7 heads is $0.215$. Find $p$.

$P(X = 7) = \binom{10}{7} p^7 (1 - p)^3 = 120 p^7 (1 - p)^3 = 0.215$.

This is a one-variable equation in $p$. Use CAS solve, restricting to $p \in (0, 1)$: $p \approx 0.6$.

Examples in context

Example 1. Germination rate of seedlings. A nursery plants $6$ seeds, each germinating independently with probability $p = 0.8$. Let $X$ be the number that germinate, so $X \sim \mathrm{Bi}(6, 0.8)$. The probability that all six germinate is $P(X = 6) = (0.8)^6 = 0.2621$. The expected number is $E(X) = 6 \times 0.8 = 4.8$ and the variance is $\mathrm{Var}(X) = 6 \times 0.8 \times 0.2 = 0.96$.

Example 2. Defective components. A batch of electronic components has a $3\%$ defect rate. In a random sample of $15$, let $X$ be the number defective, $X \sim \mathrm{Bi}(15, 0.03)$. The probability of no defectives is $P(X = 0) = (0.97)^{15} = 0.6333$, so the probability of at least one defective is $1 - 0.6333 = 0.3667$.

Try this

Q1. For $X \sim \mathrm{Bi}(8, 0.5)$, find $P(X = 4)$. [2 marks]

• Cue. $\binom{8}{4}(0.5)^4(0.5)^4 = 70 \times \frac{1}{256} = \frac{70}{256} = 0.2734$.

Q2. A goalkeeper saves $20\%$ of shots. In $10$ shots, find $E(X)$ and $\mathrm{Var}(X)$ for the number saved. [2 marks]

• Cue. $E(X) = 10 \times 0.2 = 2$; $\mathrm{Var}(X) = 10 \times 0.2 \times 0.8 = 1.6$.

Q3. A quiz has $12$ true/false questions answered by guessing. Find the probability of at least $10$ correct. [3 marks]

• Cue. $X \sim \mathrm{Bi}(12, 0.5)$; $P(X \ge 10) = \frac{\binom{12}{10} + \binom{12}{11} + \binom{12}{12}}{2^{12}} = \frac{66 + 12 + 1}{4096} = \frac{79}{4096} \approx 0.0193$.