Systematic Enumeration and Diagrams in Probability

Systematic Enumeration and Sample Spaces

To calculate probability, one must first identify the sample space, which is the set of all possible outcomes of an experiment. Systematic enumeration means listing these outcomes in a logical, ordered way. For a single event, like rolling a die, this is trivial: the outcomes are $1, 2, 3, 4, 5, 6$. However, when multiple events occur, the number of outcomes increases, and an unorganised list often leads to errors. A student must be able to count the total number of outcomes and the number of successful outcomes to find the probability $P(A) = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}}$.

Tables and Grids

Tables and grids are the most effective tools for enumerating outcomes when two independent events happen simultaneously or in succession. For example, consider rolling two fair six-sided dice and summing the scores. A $6 \times 6$ grid allows you to visualise every combination.

Example: Rolling Two Dice

If you roll two dice, there are $6 \times 6 = 36$ possible outcomes. You can represent these in a grid where the rows represent the first die and the columns represent the second die. If the question asks for the probability that the sum of the two dice is 7, you can systematically identify the outcomes in the grid: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$. Since there are 6 such outcomes, the probability is $\frac{6}{36} = \frac{1}{6}$.

Grids are particularly useful for finding the probability of regions, such as the probability that the sum is greater than 9. By looking at the bottom right of a sum-grid, you can see the outcomes $(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)$, giving a probability of $\frac{6}{36}$.

Venn Diagrams and Set Combinations

Venn diagrams are used to represent relationships between different sets or groups within a population. Although the TMUA does not require formal set notation, you must understand the logical combinations of sets. A typical Venn diagram consists of overlapping circles within a rectangle, where the rectangle represents the universal set (all possible outcomes).

1. The Overlap (A and B): This region represents outcomes that belong to both set $A$ and set $B$. In a probability context, this corresponds to the intersection.
2. The Union (A or B): This region includes everything inside circle $A$, circle $B$, or both. Note that in mathematics, 'or' is inclusive, meaning it includes the overlap.
3. The Complement (Not A): This is everything outside circle $A$. The probability of 'not $A$' is always $1 - P(A)$.

Example: Three Sets

In a class of 30 students, 15 study Maths, 12 study Physics, and 10 study Chemistry. Some students study multiple subjects: 5 study Maths and Physics, 4 study Maths and Chemistry, 3 study Physics and Chemistry, and 2 study all three. To find how many study only Maths, you must start from the centre of the Venn diagram (all three) and work outwards. Students studying Maths and Physics but not Chemistry are $5 - 2 = 3$. Students studying Maths and Chemistry but not Physics are $4 - 2 = 2$. Therefore, students studying only Maths are $15 - 3 - 2 - 2 = 8$.

Tree Diagrams

Tree diagrams are essential for sequential events where the outcome of one stage may affect the next. They are particularly useful for conditional probability problems, often described as drawing items from a bag without replacement.

Rules for Tree Diagrams:

1. Multiply along the branches: To find the probability of a specific sequence of outcomes (e.g., Red then Blue), multiply the probabilities on the corresponding branches.
2. Add the outcomes at the ends: To find the probability of a compound event (e.g., one Red and one Blue in any order), add the final probabilities of all paths that satisfy the condition.

Example: Drawing without Replacement

A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. The first branch shows $P(\text{Red}) = \frac{5}{8}$ and $P(\text{Blue}) = \frac{3}{8}$. If a red ball was drawn first, the remaining balls are 4 red and 3 blue, so the second set of branches shows $P(\text{Red}) = \frac{4}{7}$ and $P(\text{Blue}) = \frac{3}{7}$. The probability of drawing one of each colour is $P(\text{Red, Blue}) + P(\text{Blue, Red}) = (\frac{5}{8} \times \frac{3}{7}) + (\frac{3}{8} \times \frac{5}{7}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$.