AxoPrep

Mathematics: Applications and Interpretation

SL 4.6 Combined events and probability diagrams

Probabilities of combined events with Venn diagrams, tree diagrams and tables, including conditional probability and independence. Study guide, 2 practice questions with markschemes and 3 flashcards. Free.

Venn diagrams · Tree diagrams and sample-space tables · Union and intersection rule · Mutually exclusive events · Conditional probability · Independent events

Practise all 2 questions Study 3 flashcards

Study guide

Combining events

For two events \(A\) and \(B\):

Adding \(P(A)\) and \(P(B)\) counts the overlap twice, so

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Mutually exclusive events cannot happen together: \(P(A \cap B) = 0\), so \(P(A \cup B) = P(A) + P(B)\).

Conditional probability

"Given that \(B\) has happened" shrinks the sample space to \(B\):

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

In a Venn diagram or table, this is simply the count in the overlap divided by the count in the given region.

Independence

\(A\) and \(B\) are independent if one happening does not change the chance of the other: \(P(A \mid B) = P(A)\), which is equivalent to

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

To test independence, compute both sides and compare. Do not confuse independent with mutually exclusive: two events with non-zero probabilities that are mutually exclusive are never independent, because one happening makes the other impossible.

Worked example 1: Venn diagram

In a group of 80 students, 45 play football, 30 play basketball and 12 play both.

Worked example 2: tree diagram

A factory finds that 3% of phones are faulty. A test flags 95% of faulty phones and wrongly flags 2% of good ones.

So only about 60% of flagged phones are actually faulty, a surprising and very examinable result.

Choosing a diagram

Common mistakes

Exam tips

Cheat sheet

Key definitions

Intersection

The event A ∩ B that both A and B occur.

Union

The event A ∪ B that A occurs, B occurs, or both.

Mutually exclusive events

Events that cannot occur at the same time, so P(A ∩ B) = 0.

Conditional probability

The probability of A given that B has occurred, P(A | B) = P(A ∩ B) / P(B).

Independent events

Events where the occurrence of one does not affect the probability of the other; P(A ∩ B) = P(A)P(B).

Tree diagram

A branching diagram for multi-stage experiments; multiply along branches and add between them.

Practice questions

SLPaper 1Challenging

Question 1

In a group of 80 students, 45 play football (\(F\)), 30 play basketball (\(B\)) and 12 play both.

(a) [2]

Find the number of students who play neither sport.

Markscheme
  • 1 \(45 + 30 - 12 = 63\) play at least one
  • 1 \(17\)
(b) [1]

A student is chosen at random. Find \(P(F \cup B)\).

Markscheme
  • 1 \(\frac{63}{80}\)
(c) [2]

Find the probability that a student plays basketball, given that they play football.

Markscheme
  • 1 \(\frac{12}{45}\)
  • 1 \(\frac{4}{15}\)
(d) [3]

Determine whether the events \(F\) and \(B\) are independent. Justify your answer.

Markscheme
  • 1 \(P(F) \times P(B) = \frac{45}{80} \times \frac{30}{80} = 0.211\)
  • 1 \(P(F \cap B) = \frac{12}{80} = 0.15\)
  • 1 Not equal, so not independent
SLPaper 2Very hard

Question 2

A factory tests every phone screen it makes. 4% of screens are faulty. The test correctly flags 95% of faulty screens, but it also wrongly flags 3% of screens that are not faulty.

(a) [3]

Find the probability that a randomly chosen screen is flagged.

Markscheme
  • 1 Tree diagram or \(0.04 \times 0.95 + 0.96 \times 0.03\)
  • 1 \(0.038 + 0.0288\)
  • 1 \(0.0668\)
(b) [2]

A screen is flagged. Find the probability that it is actually faulty.

Markscheme
  • 1 \(\frac{0.038}{0.0668}\)
  • 1 \(0.569\)
(c) [1]

Comment on the usefulness of the test.

Markscheme
  • 1 About 43% of flagged screens are not faulty, so flagged screens need a second check (the test is still useful as a first filter)

Key facts

Test for independent events

\(A\) and \(B\) are independent if \(P(A \cap B) = P(A)\,P(B)\), or equivalently \(P(A \mid B) = P(A)\).

Conditional probability formula

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

Mutually exclusive vs independent

Mutually exclusive: cannot happen together, \(P(A \cap B) = 0\). Independent: one does not affect the other. Events with non-zero probabilities cannot be both.