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Bernoulli trials, sample data and simulation: VCE Math Methods Unit 2 Year 11

Syllabus dot point

“Bernoulli trials and sequences of Bernoulli trials, sample data analysis (mean, median, mode, range), simulation of random processes, and the relationship between theoretical probability and observed relative frequency”

8 min read

Quick answer

A Bernoulli trial has two outcomes (success with probability pp, failure with 1−p1 - p), and a sequence of nn independent Bernoulli trials produces the binomial probability P(exactly k successes)=(nk)pk(1−p)n−kP(\text{exactly } k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k} with expected value npnp; simulation (using random numbers in repeated trials) approximates theoretical probabilities, with the observed relative frequency converging to pp as the number of trials grows.

Jump to a section
  1. What this dot point is asking
  2. Bernoulli trials
  3. Sequences of Bernoulli trials
  4. Sample data analysis
  5. Theoretical probability vs observed relative frequency
  6. Simulation
  7. Connection to Unit 3 and Unit 4
  8. Examples in context
  9. Try this
  10. Exam-style questions

What this dot point is asking

VCAA wants you to recognise Bernoulli trials, compute the probability of sequences of Bernoulli outcomes, analyse sample data using summary statistics, and use simulation to approximate theoretical probabilities. The dot point is the bridge between Unit 1 probability and Unit 3 discrete random variables.

Bernoulli trials

A Bernoulli trial has exactly two outcomes: success (probability pp) and failure (probability 1−p1 - p).

Examples:

  • A coin flip: heads (success) or tails (failure).
  • A multiple-choice question answered randomly: correct or incorrect.
  • A free throw in basketball: in or out.
  • A manufacturing item: defective or not defective.

Sequences of Bernoulli trials

A Bernoulli sequence is nn independent Bernoulli trials with the same success probability pp.

For a specific sequence of outcomes (e.g., SSFSF for two successes then a failure then a success then a failure):

P(specific sequence)=p(number of S)(1−p)(number of F)P(\text{specific sequence}) = p^{\text{(number of S)}} (1-p)^{\text{(number of F)}}

For "exactly kk successes in nn trials" (any order):

P(exactly k successes)=(nk)pk(1−p)n−kP(\text{exactly } k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k}

This is the binomial probability formula, formalised in Unit 3.

Sample data analysis

When repeated trials are conducted, the resulting data can be summarised:

Mean
Sum of values divided by the number of values. Measures the centre.
Median
The middle value when data are ordered. Robust to outliers.
Mode
The most frequent value.
Range
Maximum minus minimum.
Standard deviation
Measures the spread (Unit 3 / 4 will formalise).

Theoretical probability vs observed relative frequency

For a Bernoulli trial with P(S)=pP(\text{S}) = p:

  • Theoretical probability. pp as defined.
  • Observed relative frequency. Number of successes in nn trials, divided by nn.

Law of large numbers (informal). As nn grows, the observed relative frequency tends to the theoretical probability.

For small nn, observed frequencies can differ substantially from pp. For large nn, the agreement is close.

Simulation

A simulation of a random process uses random numbers (from a calculator or coin/dice) to approximate theoretical probabilities by repeated trials.

Procedure

  1. Define the model. What is being simulated (a Bernoulli trial with pp specified)?
  2. Choose a random source. Calculator RAND, coin, dice.
  3. Decide the success condition. "If RAND <p< p, count as success."
  4. Run nn trials. Record successes.
  5. Compute the observed relative frequency. p^=\hat p = (successes) /n/ n.
  6. Compare to theoretical. For large nn, p^≈p\hat p \approx p.

Worked example. Simulating a free throw

A basketball player has free-throw probability p=0.7p = 0.7. Simulate 100 trials.

In each trial: generate RAND. If RAND <0.7< 0.7, count as a make.

After 100 trials, the observed make rate should be close to 0.70. (It would be approximately normally distributed around 0.70 with standard deviation 0.7×0.3/100=0.046\sqrt{0.7 \times 0.3 / 100} = 0.046, so most observed rates would be in the range 0.61 to 0.79.)

Why simulation matters

Simulation is the practical approach when:

  • The theoretical probability is hard to compute.
  • The system has many variables interacting.
  • You want to estimate the probability empirically without analytical work.

Modern statistical practice uses simulation heavily (Monte Carlo methods, bootstrap inference). VCE Methods introduces the concept at Year 11 level.

Connection to Unit 3 and Unit 4

Unit 3 will formalise the binomial distribution and its mean, variance and standard deviation. Unit 4 will introduce continuous random variables, the normal distribution, sample proportions and confidence intervals.

The Unit 2 foundation is the Bernoulli trial structure, the binomial probability formula (without the formal label), and the empirical vs theoretical distinction.

Common errors
Treating non-independent trials as independent
Bernoulli sequences assume each trial is independent. Drawing without replacement, where the sample changes, is not a Bernoulli sequence.
Confusing pp and 1−p1 - p
pp is the success probability; 1−p1 - p is the failure probability. Both appear in the binomial formula.
Missing the binomial coefficient
P(exactly k successes)=(nk)pk(1−p)n−kP(\text{exactly } k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k}. The (nk)\binom{n}{k} counts arrangements; without it you have only one specific sequence.
Confusing observed and theoretical
pp is fixed (theoretical). p^\hat p is observed (from data) and varies sample to sample.
Simulation with too few trials
Small nn gives unreliable estimates. The agreement improves as n\sqrt{n}.

Examples in context

Example 1. Penalty shootout sequence. A striker scores a penalty with probability p=0.8p = 0.8. In a sequence of 44 independent attempts, the probability of the specific outcome score-miss-score-score is 0.8×0.2×0.8×0.8=0.10240.8 \times 0.2 \times 0.8 \times 0.8 = 0.1024. The probability of exactly 33 goals in any order is (43)(0.8)3(0.2)=4×0.512×0.2=0.4096\binom{4}{3}(0.8)^3(0.2) = 4 \times 0.512 \times 0.2 = 0.4096, and the expected number of goals is np=4×0.8=3.2np = 4 \times 0.8 = 3.2.

Example 2. Simulating a survey response. Pollsters model a "yes" response as a Bernoulli trial with p=0.45p = 0.45. Using a calculator, each trial draws a random number in [0,1)[0, 1) and counts "yes" if it is below 0.450.45. Over 200200 simulated respondents the observed proportion p^\hat{p} clusters near 0.450.45 with spread about 0.45×0.55200≈0.035\sqrt{\tfrac{0.45 \times 0.55}{200}} \approx 0.035, so most simulation runs land between 0.380.38 and 0.520.52.

Try this

Q1. A fair die is rolled; "success" is rolling a 66. In 33 rolls, find the probability of exactly one 66. [3 marks]

  • Cue. (31)(16)(56)2=3×16×2536=75216≈0.347\binom{3}{1}\left(\tfrac16\right)\left(\tfrac56\right)^2 = 3 \times \tfrac16 \times \tfrac{25}{36} = \tfrac{75}{216} \approx 0.347.

Q2. A seed germinates with probability 0.90.9. Find the expected number that germinate from 2020 seeds. [2 marks]

  • Cue. np=20×0.9=18np = 20 \times 0.9 = 18.

Q3. Explain how observed relative frequency relates to theoretical probability as the number of trials increases. [2 marks]

  • Cue. By the law of large numbers, p^→p\hat p \to p as nn grows; small samples can deviate, large samples converge.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks
A biased coin lands heads with probability p=0.3p = 0.3. (a) For a sequence of 5 Bernoulli trials, find the probability of exactly 2 heads. (b) Estimate the expected number of heads.
Show worked answer →

(a) Probability of exactly 2 heads. Each sequence has probability p2(1−p)3=0.32×0.73=0.09×0.343=0.03087p^2 (1-p)^3 = 0.3^2 \times 0.7^3 = 0.09 \times 0.343 = 0.03087.

Number of arrangements: (52)=10\binom{5}{2} = 10.

Total: 10×0.03087≈0.30910 \times 0.03087 \approx 0.309.

(b) Expected heads. For nn Bernoulli trials with success probability pp, expected successes =np=5×0.3=1.5= n p = 5 \times 0.3 = 1.5.

Markers reward the binomial structure (Unit 3 will formalise this as the binomial distribution) and the npn p expected-value rule.

Practise this

Sources & how we know this

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