Year 12 Methods (Unit 3 & 4) Refresher on probability and discrete random variables

Conditional probability and independence

Q: What is conditional probability?

Conditional probability is the probability of one event given that another has already happened. It is written P(A given B) and calculated as P(A and B) divided by P(B). It rescales the sample space so that only outcomes in B are considered.

Q: What is the multiplication rule for probability?

The multiplication rule says P(A and B) equals P(B) times P(A given B), which also equals P(A) times P(B given A). It lets you find the probability of two events both happening by multiplying along the branches of a tree diagram.

Q: How do you test whether two events are independent?

Two events A and B are independent when P(A and B) equals P(A) times P(B), or equivalently when P(A given B) equals P(A). Compute both sides: if they are equal the events are independent, and if they differ they are dependent.

Q: What is the difference between drawing with and without replacement?

With replacement, the item is returned before the next draw, so the probabilities stay the same and the draws are independent. Without replacement, the item is not returned, so the totals fall by one and the second-stage probabilities change, making the draws dependent.

Q: How do you read a conditional probability from a two-way table?

For P(A given B), restrict attention to the row or column for B, then divide the count in both A and B by the total for B. For example if 24 of 30 cat owners also own a dog, then P(dog given cat) equals 24 divided by 30, which is 0.8.

Q: Does independent mean mutually exclusive?

No. Mutually exclusive events cannot both happen, so P(A and B) equals 0. Independent events can both happen; knowing one does not change the probability of the other, so P(A and B) equals P(A) times P(B). Two events with positive probability are never both mutually exclusive and independent.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), a conditional probability is the chance of one event given that another has occurred, \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). Rearranging gives the multiplication rule \(P(A\cap B)=P(A\mid B)\,P(B)\), the engine behind tree diagrams for two-stage experiments. Two events are independent when \(P(A\cap B)=P(A)\,P(B)\), so knowing one tells you nothing about the other — this refreshes the Units 1&2 probability you carry into Year 12.

A conditional probability \(P(A\mid B)\) is the probability that event \(A\) occurs given that event \(B\) has already occurred. Knowing \(B\) has happened shrinks the sample space to just the outcomes in \(B\):

\(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)},\qquad P(B)\neq 0.\)

Rearranging gives the multiplication rule \(P(A\cap B)=P(A\mid B)\,P(B)=P(B\mid A)\,P(A)\), which is exactly what you do when you multiply along the branches of a tree diagram for a two-stage experiment.

Events \(A\) and \(B\) are independent when the occurrence of one does not change the probability of the other. Formally, \(A\) is independent of \(B\) when \(P(A\mid B)=P(A)\), which is equivalent to \(P(A\cap B)=P(A)\,P(B)\). If this equality fails the events are dependent. Drawing with replacement keeps the draws independent; drawing without replacement makes them dependent.

Key idea. Conditional: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). Multiplication rule: \(P(A\cap B)=P(A\mid B)\,P(B)\). Independent \(\Leftrightarrow P(A\cap B)=P(A)\,P(B)\Leftrightarrow P(A\mid B)=P(A)\).

Tree without replacement (3 red, 2 green): the second-stage denominators drop from \(5\) to \(4\)

Conditioning on \(B\): of the \(24+6=30\) in \(B\), \(24\) are also in \(A\), so \(P(A\mid B)=\dfrac{24}{30}=0.8\)

Conditional probability (\(P(B)\neq 0\)):

\[P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\]

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}

The multiplication rule (rearranging the above):

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

P (A \cap B) = P (A ∣ B) P (B)

Independence of \(A\) and \(B\) (each equivalent):

\[P(A\cap B)=P(A)\,P(B)\quad\Longleftrightarrow\quad P(A\mid B)=P(A)\]

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

The complement and addition (union) rules that support these:

\[P(\overline{A})=1-P(A),\qquad P(A\cup B)=P(A)+P(B)-P(A\cap B)\]

Independent complements. If \(A\) and \(B\) are independent, then so are \(A\) and \(B'\), \(A'\) and \(B\), and \(A'\) and \(B'\). For example \(P(A'\cap B')=P(A')\,P(B')\).

How to work with conditional probability and independence

Find a conditional probability. Use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). From a two-way table, divide the count in both \(A\) and \(B\) by the total for \(B\).
Combine two events. Apply the multiplication rule \(P(A\cap B)=P(A\mid B)\,P(B)\), multiplying along the branches of a tree diagram.
Handle a two-stage experiment. Draw a tree. For without replacement, reduce the total (and the count of the item drawn) on the second stage; for with replacement, the second-stage probabilities are unchanged.
Test independence. Compute \(P(A)\,P(B)\) and compare with \(P(A\cap B)\). Equal \(\Rightarrow\) independent; not equal \(\Rightarrow\) dependent.

Language of conditionality. Phrases such as “given that”, “if”, “knowing that” and “among those who” all signal a conditional probability — the condition tells you which reduced sample space to divide by.

Example 1 — From the formula

Given \(P(A\cap B)=0.3\) and \(P(B)=0.4\), find \(P(A\mid B)\).

Solution

Divide the joint probability by \(P(B)\).

\(P(A\mid B)\)	\(=\)	\(\dfrac{P(A\cap B)}{P(B)}\)
	\(=\)	\(\dfrac{0.3}{0.4}=0.75\)

P (A ∣ B) = 0.75

Example 2 — Two-way table

Of \(100\) households, the table shows dog ownership (\(A\)) and cat ownership (\(B\)). Find \(P(A\mid B)\).

	\(A\)	\(A'\)	Total
\(B\)	24	6	30
\(B'\)	16	54	70
Total	40	60	100

Solution

Restrict to the \(B\) row: \(30\) households, \(24\) also own a dog.

\(P(A\mid B)\)

\(=\)

\(\dfrac{24}{30}=0.8\)

P (A ∣ B) = 0.8

Example 3 — Test independence

Events \(A\) and \(B\) have \(P(A)=0.6\), \(P(B)=0.5\), \(P(A\cap B)=0.2\). Are they independent?

Solution

Compare \(P(A)\,P(B)\) with \(P(A\cap B)\).

\(P(A)\,P(B)\)	\(=\)	\(0.6\times 0.5=0.30\)
\(P(A\cap B)\)	\(=\)	\(0.20\)

Since \(0.30\neq 0.20\), the events are not independent.

0.30 \neq 0.20

Example 4 — Tree, without replacement

A bag has \(5\) red and \(3\) blue counters. Two are drawn without replacement. Find \(P(\text{both red})\).

Solution

Multiply along the red–red branch; the total drops from \(8\) to \(7\).

\(P(\text{RR})\)	\(=\)	\(\dfrac{5}{8}\times\dfrac{4}{7}\)
	\(=\)	\(\dfrac{20}{56}=\dfrac{5}{14}\)

P (RR) = \frac{5}{14}

Common pitfalls

\(P(A\mid B)\) is not \(P(A\cap B)\). The conditional divides by \(P(B)\); the intersection does not. Mixing them up is the single most common error — always divide by the probability of the given event.

Do not assume independence. You may only multiply \(P(A)\,P(B)\) to get \(P(A\cap B)\) after checking the events are independent. Otherwise use \(P(A\cap B)=P(A\mid B)\,P(B)\).

Reduce the denominator without replacement. On the second draw the total is one smaller, and the count of the item already drawn is one smaller too. Forgetting this treats the draws as if they were with replacement.

Frequently asked questions

What is conditional probability?

The probability of \(A\) given that \(B\) has occurred, \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). Knowing \(B\) rescales the sample space to the outcomes in \(B\).

What is the multiplication rule for probability?

\(P(A\cap B)=P(A\mid B)\,P(B)=P(B\mid A)\,P(A)\). It is what you use when multiplying along the branches of a tree diagram.

How do you test whether two events are independent?

Check whether \(P(A\cap B)=P(A)\,P(B)\) (equivalently \(P(A\mid B)=P(A)\)). Equal means independent; not equal means dependent.

What is the difference between drawing with and without replacement?

With replacement the item is returned, so the draws are independent and probabilities stay fixed. Without replacement the totals fall by one, so the draws are dependent.

How do you read a conditional probability from a two-way table?

Restrict to the row or column for the given event, then divide the joint count by that event's total. E.g. \(24\) of \(30\) cat owners own a dog gives \(P(A\mid B)=0.8\).

Does independent mean mutually exclusive?

No. Mutually exclusive means \(P(A\cap B)=0\); independent means \(P(A\cap B)=P(A)\,P(B)\). Events with positive probability cannot be both.

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