1. Overview
Probability of combined events deals with calculating the likelihood of two or more events occurring together or in sequence. This involves understanding how to represent all possible outcomes and applying appropriate rules for calculating probabilities, especially when events are independent, mutually exclusive, or dependent. Sample space diagrams, Venn diagrams, and tree diagrams are essential tools for solving these problems.
Key Definitions
- Combined Event: An outcome that depends on two or more separate events (e.g., rolling a die AND flipping a coin).
- Sample Space: The set of all possible outcomes of an experiment.
- Independent Events: Events where the outcome of one does not affect the probability of the other.
- Mutually Exclusive: Events that cannot happen at the same time (e.g., a card cannot be both a Heart and a Spade).
- Exhaustive: A set of events that covers all possible outcomes (their probabilities sum to 1).
Core Content
A. Sample Space Diagrams
Sample space diagrams are usually grids used to show all possible outcomes when two independent events are combined (e.g., rolling two dice).
Worked example 1 — Rolling two dice
Two fair four-sided dice, numbered 1 to 4, are rolled. The scores are added together. Find the probability that the total score is 5.
Step 1: Draw the sample space grid.
| + | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 |
| 2 | 3 | 4 | 5 | 6 |
| 3 | 4 | 5 | 6 | 7 |
| 4 | 5 | 6 | 7 | 8 |
Step 2: Count the total outcomes. Total outcomes = $4 \times 4 = 16$. Each die has 4 possible outcomes, so the total number of combined outcomes is 4 multiplied by 4.
Step 3: Count the successful outcomes (where sum = 5). The sums that equal 5 are: (1,4), (2,3), (3,2), and (4,1). There are 4 such outcomes.
Step 4: Calculate the probability. $P(\text{sum is 5}) = \frac{4}{16}$ Probability is the number of successful outcomes divided by the total number of outcomes.
$P(\text{sum is 5}) = \frac{1}{4}$
$P(\text{sum is 5}) = 0.25$
Answer: The probability that the total score is 5 is $\boxed{\frac{1}{4}}$ or $\boxed{0.25}$.
Worked example 2 — Spinner and Coin
A spinner has three sections, labelled A, B, and C. The spinner is spun and a fair coin is flipped. Draw a sample space diagram to show all possible outcomes. What is the probability of getting section A on the spinner and a tail on the coin?
Step 1: Draw the sample space grid. The horizontal axis represents the spinner (A, B, C) and the vertical axis represents the coin (H, T).
| A | B | C | |
|---|---|---|---|
| H | A, H | B, H | C, H |
| T | A, T | B, T | C, T |
Step 2: Count the total outcomes. Total outcomes = $3 \times 2 = 6$ There are 3 possible outcomes for the spinner and 2 for the coin.
Step 3: Count the successful outcomes (A and T). There is only one outcome where the spinner lands on A and the coin lands on tails: (A, T).
Step 4: Calculate the probability. $P(\text{A and T}) = \frac{1}{6}$ Probability is the number of successful outcomes divided by the total number of outcomes.
Answer: The probability of getting section A on the spinner and a tail on the coin is $\boxed{\frac{1}{6}}$.
B. Venn Diagrams
Venn diagrams show the relationship between different sets of outcomes. They are useful for "And" ($\cap$) and "Or" ($\cup$) probabilities.
Worked example 1 — Sports in a class
In a class of 30 students, 18 play Football (F), 15 play Cricket (C), and 8 play both. Find the probability that a student chosen at random plays neither sport.
Step 1: Fill in the intersection first. $P(F \cap C) = 8$. The intersection represents students who play both sports.
Step 2: Calculate the remaining sections.
- Football only: $18 - 8 = 10$.
- Cricket only: $15 - 8 = 7$. Subtract the number who play both from the total number who play each sport to find those who play only that sport.
Step 3: Find the number of students who play neither. Total students = 30. Neither = $30 - (10 + 8 + 7)$ Subtract the number of students who play at least one sport from the total number of students.
Neither = $30 - 25 = 5$.
Step 4: Calculate probability. $P(\text{Neither}) = \frac{5}{30}$ Probability is the number of students who play neither sport divided by the total number of students.
$P(\text{Neither}) = \frac{1}{6}$.
Answer: The probability that a student chosen at random plays neither sport is $\boxed{\frac{1}{6}}$.
Worked example 2 — Sets and Venn Diagrams
In a group of 50 people, 30 like coffee, 25 like tea, and 10 like neither. What is the probability that a person chosen at random likes both coffee and tea?
Step 1: Define the sets. Let C be the set of people who like coffee. Let T be the set of people who like tea.
Step 2: Find the number of people who like at least one drink. People who like at least one drink = Total people - People who like neither People who like at least one drink = $50 - 10 = 40$
Step 3: Use the formula for the union of two sets. $n(C \cup T) = n(C) + n(T) - n(C \cap T)$ Where: $n(C \cup T)$ = Number of people who like coffee or tea or both = 40 $n(C)$ = Number of people who like coffee = 30 $n(T)$ = Number of people who like tea = 25 $n(C \cap T)$ = Number of people who like both coffee and tea (what we want to find)
Step 4: Substitute the values and solve for $n(C \cap T)$. $40 = 30 + 25 - n(C \cap T)$ $40 = 55 - n(C \cap T)$ $n(C \cap T) = 55 - 40$ $n(C \cap T) = 15$
Step 5: Calculate the probability. $P(\text{Both}) = \frac{n(C \cap T)}{\text{Total people}}$ $P(\text{Both}) = \frac{15}{50}$ $P(\text{Both}) = \frac{3}{10}$
Answer: The probability that a person chosen at random likes both coffee and tea is $\boxed{\frac{3}{10}}$.
C. Tree Diagrams
Tree diagrams are used for sequential events.
- Multiply probabilities along the branches (to find "Event A AND Event B").
- Add the probabilities at the ends of the branches (to find "Outcome 1 OR Outcome 2").
Worked example 1 — Beads in a bag (with replacement)
A bag contains 5 Red beads and 3 Blue beads. A bead is picked at random, the color is recorded, and then it is replaced. A second bead is then picked. Find the probability of picking one bead of each color.
Step 1: Calculate individual probabilities. $P(\text{Red}) = \frac{5}{8}$ There are 5 red beads out of a total of 8. $P(\text{Blue}) = \frac{3}{8}$ There are 3 blue beads out of a total of 8.
Step 2: Draw the tree diagram.
Step 3: Identify the "one of each color" paths. Path 1: Red then Blue $\rightarrow P(R, B) = \frac{5}{8} \times \frac{3}{8}$ Multiply the probabilities along the branches. $P(R, B) = \frac{15}{64}$ Path 2: Blue then Red $\rightarrow P(B, R) = \frac{3}{8} \times \frac{5}{8}$ Multiply the probabilities along the branches. $P(B, R) = \frac{15}{64}$
Step 4: Add the results. $P(\text{one of each}) = \frac{15}{64} + \frac{15}{64}$ Add the probabilities of the two paths that result in one of each color. $P(\text{one of each}) = \frac{30}{64}$ $P(\text{one of each}) = \frac{15}{32}$.
Answer: The probability of picking one bead of each color is $\boxed{\frac{15}{32}}$.
Worked example 2 — Beads in a bag (without replacement)
A bag contains 4 green balls and 6 yellow balls. A ball is selected at random and not replaced. A second ball is then selected. Calculate the probability that the first ball is green and the second ball is yellow.
Step 1: Calculate the probability of the first ball being green. $P(\text{First ball is Green}) = \frac{\text{Number of green balls}}{\text{Total number of balls}} = \frac{4}{10} = \frac{2}{5}$
Step 2: Calculate the probability of the second ball being yellow, given the first was green and not replaced. Since the first ball was green and not replaced, there are now 3 green balls and 6 yellow balls left in the bag. The total number of balls is now 9. $P(\text{Second ball is Yellow | First ball is Green}) = \frac{\text{Number of yellow balls}}{\text{Total number of balls remaining}} = \frac{6}{9} = \frac{2}{3}$
Step 3: Calculate the probability of both events occurring. $P(\text{First ball is Green AND Second ball is Yellow}) = P(\text{First ball is Green}) \times P(\text{Second ball is Yellow | First ball is Green})$ $P(\text{Green then Yellow}) = \frac{2}{5} \times \frac{2}{3}$ $P(\text{Green then Yellow}) = \frac{4}{15}$
Answer: The probability that the first ball is green and the second ball is yellow is $\boxed{\frac{4}{15}}$.
Extended Content (Extended Only)
Note: This specific topic (8.3) is designated as Core curriculum. More advanced conditional probability is covered in 8.4. While topic 8.3 is Core, it's important to deepen understanding of the core concepts for students aiming for higher grades. This involves tackling more complex scenarios within sample space diagrams, Venn diagrams, and tree diagrams. For example, with Venn diagrams, problems might involve three overlapping sets instead of two, requiring careful calculation of each region. With tree diagrams, questions might involve multiple stages or require working backwards to find a missing probability. These more complex problems build a stronger foundation for tackling conditional probability later on.
Key Equations
- General Probability: $\qquad P(A) = \frac{\text{Number of ways A can occur}}{\text{Total number of outcomes}}$
- The "AND" Rule (Independent): $\qquad P(A \text{ and } B) = P(A) \times P(B)$
- The "OR" Rule (Mutually Exclusive): $\qquad P(A \text{ or } B) = P(A) + P(B)$
- Complementary Events: $\qquad P(\text{Not } A) = 1 - P(A)$
Common Mistakes to Avoid
- ❌ Adding along branches: Students often add probabilities when moving across a tree diagram instead of multiplying.
- ✓ Right: Always multiply probabilities along the branches to find the probability of a sequence of events.
- ❌ Forgetting "Without Replacement": Students use the same denominator for the second event even when an item wasn't put back. This changes the probabilities for subsequent events.
- ✓ Right: If an item is not replaced, the denominator (total) and numerator (if the same color) must decrease by 1. Recalculate the probabilities for each stage of the tree diagram.
- ❌ Missing paths: Only calculating Red-Blue and forgetting Blue-Red when the question asks for the probability of getting one of each color, regardless of order.
- ✓ Right: Read the question carefully to see if order matters. If it says "one of each," you must account for all possible orders and add their probabilities.
- ❌ Incorrectly filling Venn Diagrams: Not starting with the intersection of sets when using Venn diagrams.
- ✓ Right: Always fill in the intersection (the "and" part) first, then work outwards to find the probabilities of each individual section.
Exam Tips
- Calculator vs Non-Calculator: If a question uses fractions, stay in fractions to avoid rounding errors. If it uses decimals, your calculator is your best friend—just ensure you don't round too early in the calculation.
- Command Words: "Complete the tree diagram" usually awards 1-2 marks for the branches and probabilities. "Show that..." means you must write down every step of the multiplication and addition to reach the given answer.
- The "Sum to 1" Check: Always check that the branches coming from a single point on a tree diagram add up to exactly 1. If they don't, you've made an arithmetic error.
- Real-world Context: Expect questions about weather (Rain/No Rain), passing driving tests, or picking different colored sweets from a jar.
- Notation: Use $P(A)$ notation in your working to keep it clear for the examiner. This helps you get "method marks" even if your final answer is wrong.