Bernoulli trials | Year 12

01 / LEARN

Understand the idea.

Success is a label, not necessarily a good result

A Bernoulli trial sorts its outcome into success or failure. Success is simply the event being counted, even if it is a fault. Define an indicator $B = 1$ for that event and $B = 0$ otherwise.

If success has probability $p$ , then $P (B = 1) = p$ and $P (B = 0) = 1 - p$ . One indicator records whether an event occurred; it does not record how many times the event occurred within a larger count.

Derive, rather than memorise, the summaries

Derive the mean by weighting its two values: $E (B) = 0 (1 - p) + 1 p = p$ . Since $0^{2} = 0$ and $1^{2} = 1$ , the same calculation gives $E (B^{2}) = p$ , hence $Var (B) = p (1 - p)$ .

At $p = 0$ or $p = 1$ the indicator is certain and its variance is zero. A payoff such as $5 B$ uses different outcome values, so it is a scaled indicator; apply the scale rules to find its moments.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An event called success

A sensor passes a check with probability 0.65. Define B = 1 for a pass. Find its mean and variance.

Worked solution

1
Record the two outcomes.
$P (B = 1) = 0.65, P (B = 0) = 0.35$
The indicator equals one for a pass and zero for its complement.
2
Average the indicator.
$E (B) = 0 (0.35) + 1 (0.65) = 0.65$
The mean of an indicator equals the probability of the event it records.
3
Compute its variance.
$Var (B) = 0.65 (1 - 0.65) = 0.65 (0.35) = 0.2275$
Bernoulli variance includes uncertainty in both possible outcomes.

Answer

Mean 0.65; variance 0.2275.

EXAMPLE 02

Turn a count event into an indicator

X is 0, 1 or 2 with probabilities 0.4, 0.35 and 0.25. Let B = 1 if X ≥ 1 and 0 otherwise. Find the distribution and variance of B.

Worked solution

1
Combine the outcomes that trigger the indicator.
$p = P (X \geq 1) = P (X = 1) + P (X = 2) = 0.35 + 0.25 = 0.60$
Both positive counts map to the same indicator value one.
2
Write the indicator distribution.
$P (B = 0) = 0.40, P (B = 1) = 0.60$
Only $X = 0$ maps to zero, so its original probability remains $0.40$ .
3
Calculate the variance.
$Var (B) = 0.60 (0.40) = 0.24$
The transformed variable is Bernoulli even though the original count had three values.

Answer

B has probabilities 0.4, 0.6 at 0, 1; variance 0.24.

03 / YOUR TURN

Try it yourself.

Check your understanding

For B Bernoulli with p = 0.2, find P(B = 0).

Show answer & reasoning

0.8.

The complement of success has probability $1 - p = 1 - 0.2 = 0.8$ , so $P (B = 0) = 0.8$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

A coin lands heads with probability 0.55. Define a Bernoulli indicator for tails and give $p$ .
2 marks
$B$ is Bernoulli with $p = 0.4$ . Find $E (B)$ , $Var (B)$ and $SD (B)$ .
3 marks
Let $B$ be Bernoulli with $p = 0.1$ . A bonus is $Y = 20 B$ dollars. Find its mean and variance.
3 marks
If $B$ is Bernoulli with $p = 1$ , describe $B$ and find its variance.
2 marks
A variable equals 2 or 7. Explain how a Bernoulli variable $B$ can represent it.
2 marks
Show that a Bernoulli variance cannot exceed $\frac{1}{4}$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A coin lands heads with probability 0.55. Define a Bernoulli indicator for tails and give $p$ .
2 marks
The chosen success event is tails.
Its probability is $1 - 0.55$ .
$B = 1$ for tails, 0 for heads; $p = 0.45$ .
$B$ is Bernoulli with $p = 0.4$ . Find $E (B)$ , $Var (B)$ and $SD (B)$ .
3 marks
Use $E (B) = p$ .
Use $Var (B) = p (1 - p)$ and take its square root.
$0.4; 0.24; \sqrt{0.24} \approx 0.4899$ .
Let $B$ be Bernoulli with $p = 0.1$ . A bonus is $Y = 20 B$ dollars. Find its mean and variance.
3 marks
$E (Y) = 20 (0.1) = 2$ .
$Var (Y) = 20^{2} (0.1) (0.9) = 36$ .
Mean 2 dollars; variance 36 dollars².
If $B$ is Bernoulli with $p = 1$ , describe $B$ and find its variance.
2 marks
Success is certain.
The formula $p (1 - p)$ gives $1 \times 0 = 0$ .
$B$ is always 1; variance 0.
A variable equals 2 or 7. Explain how a Bernoulli variable $B$ can represent it.
2 marks
If $B = 0$ the expression is 2.
If $B = 1$ it is 7; $p$ is the probability of the outcome 7.
$X = 2 + 5 B$ , with $B = 1$ for the outcome 7.
Show that a Bernoulli variance cannot exceed $\frac{1}{4}$ .
3 marks
Complete the square in $p - p^{2}$ .
A square is non-negative, with equality in the bound when $p = \frac{1}{2}$ .
$p (1 - p) = \frac{1}{4} - {(p - \frac{1}{2})}^{2} \leq \frac{1}{4}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Success is a label, not necessarily a good result

Derive, rather than memorise, the summaries

See the method in action.

An event called success

Record the two outcomes.

Average the indicator.

Compute its variance.

Turn a count event into an indicator

Combine the outcomes that trigger the indicator.

Write the indicator distribution.

Calculate the variance.

Try it yourself.

Check your understanding

Keep practising