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Year 11 Methods (Unit 1 & 2) Probability

Independent Events

20 practice questions 1 video lesson Theory + worked examples

Understand independent events for Queensland Year 11 Mathematical Methods (QCAA). Two events are independent when one happening does not change the chance of the other, so their probabilities simply multiply together.

You will learn to test whether events are independent, multiply probabilities across separate stages, tell independence apart from mutually exclusive events, and handle "at least one" questions using the complement — a key idea for tree diagrams and everyday probability.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1 Topic 5), two events are independent when one does not affect the other, defined by \(P(A\cap B)=P(A)\,P(B)\). This page shows how to compute a joint probability by multiplying, test independence, combine it with the complement and addition rules, and handle multi-stage independent trials.

Events \(A\) and \(B\) are independent when the occurrence of one does not change the probability of the other. The test is \(P(A\cap B)=P(A)\,P(B)\); equivalently \(P(A\mid B)=P(A)\).

Independence is not the same as mutually exclusive. Mutually exclusive events cannot both occur (\(P(A\cap B)=0\)); independent events usually can, and their joint probability is the product of the separate probabilities.

For a chain of independent trials you multiply the probabilities, and ‘at least one’ is handled with the complement \(1-P(\text{none})\).

Multiply to test and to build. If \(P(A\cap B)=P(A)P(B)\) the events are independent; and for independent events you find \(P(A\cap B)\) by multiplying.
Two independent days for a machine that works with probability 0.9Because the days are independent, each branch keeps probability zero point nine; the products add to one. 0.9 W 0.9 W WW = 0.81 0.1 F WF = 0.09 0.1 F 0.9 W FW = 0.09 0.1 F FF = 0.01
With independent stages the branch probabilities do not change: \(P(\text{works both days})=0.9\times 0.9=0.81\).
Test independence: is P(A cap B)=P(A)P(B)?
\(B\)\(B'\)Total
\(A\)0.200.300.50
\(A'\)0.200.300.50
Total0.400.601
To test independence, compare the corner cell with the product of the margins: here \(P(A)P(B)=0.5\times 0.4=0.2=P(A\cap B)\), so \(A\) and \(B\) are independent.

The definition / test of independence:

\[P(A\cap B)=P(A)\,P(B)\]
P(A∩B)=P(A)P(B)

For independent events the complements are independent too:

\[P(A'\cap B')=P(A')\,P(B')\]
P(A′∩B′)=P(A′)P(B′)

At least one over independent trials:

\[P(\text{at least one})=1-P(\text{none})\]
P(at least one)=1-P(none)
Independent vs mutually exclusive: independent means \(P(A\cap B)=P(A)P(B)\); mutually exclusive means \(P(A\cap B)=0\). They are different ideas.

How to work with independent events

  1. Compute a joint probability: for independent events multiply, \(P(A\cap B)=P(A)P(B)\).
  2. Test independence: check whether \(P(A\cap B)\) equals \(P(A)P(B)\).
  3. Combine rules: use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) and independent complements \(P(A'\cap B')=P(A')P(B')\).
  4. Multi-stage: multiply along the independent trials; use \(1-P(\text{none})\) for ‘at least one’.
Example 1 — Multiplying independent probabilities
Events \(A\) and \(B\) are independent with \(P(A)=0.3\) and \(P(B)=0.6\). Find \(P(A\cap B)\).
Solution

For independent events, multiply:

\(P(A\cap B)\)\(=\)\(P(A)\,P(B)\)
\(=\)\(0.3\times 0.6\)

Multiply:

\(=\)\(0.18\)

\(P(A\cap B)=0.18\).

Product of two independent probabilitiesA only zero point one two, both zero point one eight, B only zero point four two, neither zero point two eight. A B 0.12 0.18 0.42 0.28
Example 2 — Testing independence
For events \(A\) and \(B\), \(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.2\). Are \(A\) and \(B\) independent?
Two-way probability table
\(B\)\(B'\)Total
\(A\)0.200.300.50
\(A'\)0.200.300.50
Total0.400.601
Solution

Work out the product of the separate probabilities:

\(P(A)\,P(B)\)\(=\)\(0.5\times 0.4\)
\(=\)\(0.2\)

Compare with the given joint probability:

\(P(A\cap B)\)\(=\)\(0.2\)

Since \(P(A\cap B)=P(A)P(B)=0.2\), the test is satisfied.

Yes — \(A\) and \(B\) are independent, because \(P(A\cap B)=P(A)P(B)\).

Example 3 — Union and neither (independent)
With the same independent events (\(P(A)=0.5\), \(P(B)=0.4\), \(P(A\cap B)=0.2\)), find \(P(A\cup B)\) and \(P(A'\cap B')\).
Solution

Addition rule for the union:

\(P(A\cup B)\)\(=\)\(0.5+0.4-0.2\)
\(=\)\(0.7\)

Neither, using independent complements:

\(P(A'\cap B')\)\(=\)\(P(A')\,P(B')\)
\(=\)\((1-0.5)(1-0.4)\)
\(=\)\(0.5\times 0.6\)
\(=\)\(0.3\)

\(P(A\cup B)=0.7\) and \(P(A'\cap B')=0.3\).

Neither of two independent eventsA only zero point three, both zero point two, B only zero point two; the shaded neither region is zero point three. A B 0.30 0.20 0.20 0.30
Example 4 — Multi-stage independent trials
A machine works on any day, independently, with probability \(0.9\). Find the probability it works on all of \(3\) days, and the probability it fails on at least one day.
Solution

Works all three days — multiply the independent probabilities:

\(P(\text{all work})\)\(=\)\(0.9\times 0.9\times 0.9\)
\(=\)\(0.9^3\)
\(=\)\(0.729\)

At least one failure is the complement:

\(P(\text{at least one fail})\)\(=\)\(1-0.729\)
\(=\)\(0.271\)

\(P(\text{all work})=0.729\) and \(P(\text{at least one fail})=0.271\).

Three independent days for the machineThe all-work path has probability zero point nine cubed, which is zero point seven two nine. 0.9 W 0.9 W 0.9 W WWW = 0.729 0.1 F WWF 0.1 F 0.9 W WFW 0.1 F WFF 0.1 F 0.9 W 0.9 W FWW 0.1 F FWF 0.1 F 0.9 W FFW 0.1 F FFF

Common pitfalls

Confusing independent with mutually exclusive. Independent means \(P(A\cap B)=P(A)P(B)\); mutually exclusive means \(P(A\cap B)=0\). Mutually exclusive events with non-zero probabilities are never independent.
Adding instead of multiplying. For an ‘and’ of independent events you multiply; addition is for a union.
Assuming independence. Only multiply \(P(A)P(B)\) when the events are stated or shown to be independent — otherwise use \(P(A\mid B)P(B)\).
Doing ‘at least one’ the long way. Use \(1-P(\text{none})\) instead of adding many separate cases.

Frequently asked questions

What does it mean for two events to be independent?

One event does not change the probability of the other. Formally \(P(A\cap B)=P(A)\,P(B)\), or equivalently \(P(A\mid B)=P(A)\).

How do you test if two events are independent?

Check whether \(P(A\cap B)=P(A)\,P(B)\). If the joint probability equals the product of the two probabilities, they are independent.

Are independent and mutually exclusive the same thing?

No. Mutually exclusive means \(P(A\cap B)=0\); independent means \(P(A\cap B)=P(A)P(B)\). They are different, and usually incompatible.

How do you find P(A and B) for independent events?

Multiply the separate probabilities: \(P(A\cap B)=P(A)\,P(B)\).

How do you find ‘at least one’ over independent trials?

Use the complement: \(P(\text{at least one})=1-P(\text{none})\), where \(P(\text{none})\) is the product of the ‘not’ probabilities.

If A and B are independent, are A' and B' independent?

Yes — \(P(A'\cap B')=P(A')\,P(B')\) as well.

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