Given P(A B) = 0.12 and P(B) = 0.3, find P(A B). [2 marks]

SEAB Original-style practice: Events A and B have P(A) = 0.6, P(B) = 0.5 and P(A B) = 0.3. Find P(A B) and determine whether A and B are independent.

P(A B) = P(A B)P(B) = 0.30.5 = 0.6. For independence, check whether P(A B) = P(A)P(B): 0.6 × 0.5 = 0.3, which equals P(A B). So A and B are independent (equivalently P(A B) = P(A)). Markers reward the conditional formula, the independence test, and a correct conclusion.

SingaporeMathsSyllabus dot point

How does conditioning on information change a probability, and what does independence mean?

Calculate conditional probabilities, test for independence, and apply the conditional probability formula and the law of total probability

A focused answer to the H2 Mathematics outcome on conditional probability. The conditional formula, testing independence, the law of total probability, and reasoning with given information.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to calculate conditional probabilities, test whether two events are independent, and use the law of total probability to combine probabilities across a partition of cases.

The answer

Conditional probability

The probability of $A$ given that $B$ has occurred is

\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}, \qquad \mathrm{P}(B) > 0.

Conditioning on $B$ restricts the sample space to outcomes in $B$ , so we rescale by $\mathrm{P}(B)$ .

Independence

Events $A$ and $B$ are independent if knowing one does not change the probability of the other. The equivalent tests:

\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) \iff \mathrm{P}(A \mid B) = \mathrm{P}(A).

If either holds, both do. Independence is a property to be tested, not assumed.

The law of total probability

If the events $B_1, B_2, \ldots$ partition the sample space (mutually exclusive and exhaustive), then

\mathrm{P}(A) = \sum_i \mathrm{P}(B_i)\mathrm{P}(A \mid B_i).

This sums the contributions to $A$ from each case, weighted by how likely each case is. It is exactly the "add across the branches of a tree" rule.

Reversing the condition

When a question asks for $\mathrm{P}(B \mid A)$ from $\mathrm{P}(A \mid B)$ , use $\mathrm{P}(B \mid A) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)}$ , computing the denominator by the law of total probability if needed.

Worked example

In a class, $70\%$ study Mathematics, $40\%$ study Physics, and $30\%$ study both. A student is chosen at random. Find the probability they study Physics given they study Mathematics, and state whether the two subjects are independent.

Step 1: Write the given probabilities

$\mathrm{P}(M) = 0.7$ , $\mathrm{P}(P) = 0.4$ , $\mathrm{P}(M \cap P) = 0.3$ .

Step 2: Apply the conditional formula

\mathrm{P}(P \mid M) = \frac{\mathrm{P}(P \cap M)}{\mathrm{P}(M)} = \frac{0.3}{0.7} = \frac{3}{7} \approx 0.429

Step 3: Test independence

Check $\mathrm{P}(M)\mathrm{P}(P) = 0.7 \times 0.4 = 0.28$ against $\mathrm{P}(M \cap P) = 0.3$ .

Step 4: Conclude

Since $0.28 \neq 0.30$ , the events are not independent: studying Mathematics slightly raises the chance of studying Physics (the conditional $0.429$ exceeds $\mathrm{P}(P) = 0.4$ ).

Examples in context

Example 1. Diagnostic testing. The chance a person has a disease given a positive test reverses the test's known sensitivity using conditional probability and the law of total probability, the calculation behind interpreting medical screening results.

Example 2. Spam filtering. An email filter updates the probability that a message is spam given a flagged word by conditioning, exactly the conditional-probability reasoning that powers simple spam detectors.

Try this

Q1. Given $\mathrm{P}(A \cap B) = 0.12$ and $\mathrm{P}(B) = 0.3$ , find $\mathrm{P}(A \mid B)$ . [2 marks]

Cue. $\dfrac{0.12}{0.3} = 0.4$ .

Q2. State the test for two events to be independent. [1 mark]

Cue. $\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$ .

Q3. Explain why mutually exclusive events with non-zero probability cannot be independent. [2 marks]

Cue. If one occurs the other cannot, so $\mathrm{P}(A \mid B) = 0 \neq \mathrm{P}(A)$ , contradicting independence.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksEvents

A

and

B

have

\mathrm{P}(A) = 0.6

\mathrm{P}(B) = 0.5

and

\mathrm{P}(A \cap B) = 0.3

. Find

\mathrm{P}(A \mid B)

and determine whether

A

and

B

are independent.

Show worked answer →

$\mathrm{P}(A \mid B) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} = \dfrac{0.3}{0.5} = 0.6$ .

For independence, check whether $\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$ : $0.6 \times 0.5 = 0.3$ , which equals $\mathrm{P}(A \cap B)$ .

So $A$ and $B$ are independent (equivalently $\mathrm{P}(A \mid B) = \mathrm{P}(A)$ ).

Markers reward the conditional formula, the independence test, and a correct conclusion.

Original5 marksMachine X makes

60\%

of items with a

2\%

defect rate; machine Y makes the rest with a

5\%

defect rate. Find the probability that a randomly chosen item is defective.

Show worked answer →

By the law of total probability:
$\mathrm{P}(\text{defective}) = \mathrm{P}(X)\mathrm{P}(\text{def} \mid X) + \mathrm{P}(Y)\mathrm{P}(\text{def} \mid Y)$ .

$= 0.6 \times 0.02 + 0.4 \times 0.05 = 0.012 + 0.020 = 0.032$ .

So the probability an item is defective is $0.032$ (that is $3.2\%$ ).

Markers reward partitioning by machine, the weighted sum, and the total probability $0.032$ .