Basic Probability (AA HL)
Before tree diagrams, conditional probability or Bayes' theorem come into play, most questions start from the same idea: counting outcomes. If every outcome in a sample space is equally likely, probability is just a ratio - favourable outcomes over total outcomes. This page covers building and reading sample spaces correctly, which is where most early marks are won or lost. It's part of the broader Probability topic.
14 questions on this sub-topic.
Counting outcomes
Covered under IB syllabus reference SL4.6: use of sample space diagrams and tables of outcomes to count equally likely results before applying any combined-event rule.
Equally likely outcomes
\[P(A)=\dfrac{n(A)}{n(U)}\]
\(n(A)\) is the number of favourable outcomes, \(n(U)\) is the total number of outcomes in the sample space. Only valid when every outcome is equally likely.
Two-event sample spaces
For two dice, two spins, or any pair of independent trials, a grid with one event along each axis lists every combined outcome exactly once - much safer than trying to count by hand.
Need the full syllabus wording and formula-booklet reference table? See Probability.
Worked examples
A fair die is rolled.
(a) Find the probability of obtaining a six.
(b) Find the probability of obtaining an even number.
Worked solution
(a) \(P(\text{six}) = \dfrac16.\) M1
\(P(\text{six}) = \dfrac16.\) A1
(b) \(P(\text{even}) = \dfrac{3}{6} = \dfrac12.\) A1
Two fair six-sided dice are rolled.
(a) Find the probability the product is 12.
(b) Find the probability the product is odd.
Worked solution
(a) Sample space \(=36\) outcomes. M1
Product 12. Pairs \((2,6),(6,2),(3,4),(4,3)\) - 4 ways: \(\dfrac{4}{36}=\dfrac19.\) A1
(b) Product odd. A product is odd only when both dice are odd M1
\(\{1,3,5\}:\) \(3\times3=9\) outcomes, \(\dfrac{9}{36}=\dfrac14.\) A1
Common mistakes
- Assuming outcomes are equally likely when they aren't. "Sum of two dice" outcomes range from 2 to 12, but they don't occur with equal probability - a sum of 7 has far more ways to occur than a sum of 2. Count outcome pairs, not sums, when in doubt.
- Missing repeated outcomes in a sample space grid. \((3,4)\) and \((4,3)\) are different outcomes even though the product or sum is the same - a grid with one die on each axis stops these being double-counted or dropped.
- Confusing "at least one" with "exactly one". "At least one six" includes rolling two sixes; the safer route is usually to find the complement, \(P(\text{no sixes})\), and subtract from 1.
Ready to practise properly?
14 basic probability questions, marked instantly like the real exam.
Quick answers
How do you find a basic probability?
For equally likely outcomes, \(P(A)=\dfrac{n(A)}{n(U)}\) - the number of favourable outcomes divided by the total number of outcomes. List or count the sample space carefully before dividing.
How do you build a sample space for two dice or two events?
List every possible pair of outcomes, usually as a grid or table, then count how many of those pairs satisfy the condition you need before dividing by the total. For calculator tips, see the Probability GDC section.