Welcome to Combined Events and Tree Diagrams
Have you ever wondered what the chances are of flipping two heads in a row, rolling a double-six in a board game, or picking two red socks from a messy drawer in the morning? In everyday life, events rarely happen all by themselves. We often want to know the probability of two or more things happening one after another or at the exact same time. In GCSE Mathematics (CCEA 2210), these are called combined events.
Don't worry if probability feels tricky right now! A tree diagram is simply a visual map that helps you organize every possible outcome step-by-step. By following a few straightforward rules, you can solve even the most complex exam questions with total confidence.
Key Takeaway: Combined events look at two or more trials happening together or in sequence. Tree diagrams turn complicated probability questions into easy-to-follow visual paths!
1. Key Definitions & Building Blocks
Before drawing diagrams, let's make sure we understand the core terms used by CCEA examiners:
• Outcome: The single result of an experiment (for example, rolling a \(4\) on a die, or flipping a Tail on a coin).
• Combined / Compound Events: Scenarios involving two or more successive or simultaneous trials (such as tossing two coins or picking two beads from a bag).
• Mutually Exclusive Events: Outcomes that cannot happen at the same time. For example, a single coin toss cannot be both Heads and Tails at once. For mutually exclusive events, we add their probabilities: \(P(A \text{ or } B) = P(A) + P(B)\).
• Independent Events: The outcome of the first event has no effect on the probability of the second event. Examples include flipping two separate coins, rolling a die twice, or picking a sweet from a bag and putting it back (with replacement).
• Dependent Events (Conditional Probability): The outcome of the first event changes the probability of the next event. The classic example is picking items from a bag without putting them back (without replacement).
Did You Know? On your CCEA papers (Units M6, M7, and M8), independent events appear widely, while dependent events ("without replacement") are a key focus in higher-tier completion units such as M8!
Key Takeaway: If the first choice changes what is left for the second choice, the events are dependent. If everything resets to the start, they are independent.
2. Anatomy of a Tree Diagram
A tree diagram grows from left to right, spreading out into "branches" for each stage of an experiment.
How a Tree Diagram is Built:
• Nodes (Branch Points): Every point where branches split represents a single trial or stage (e.g., 1st Pick, 2nd Pick).
• Branches: Each branch line represents one possible outcome of that trial. The probability is written clearly along the branch.
• The Branch Sum Rule: The probabilities on any set of branches meeting at a single node must always add up to \(1\). If one branch is \(\frac{3}{10}\), the other branch at that fork must be \(\frac{7}{10}\) because \(1 - \frac{3}{10} = \frac{7}{10}\).
• Paths: A complete route traced from the starting root on the left to the far right tip represents a full combined outcome.
• End Outcomes Column: Always list the combined outcomes at the far right end of each path (e.g., \(RR, RG, GR, GG\)).
Memory Trick: The Two Golden Rules of Tree Diagrams
To navigate any tree diagram successfully, memorize these two simple directions:
1. Multiply ACROSS branches: When moving along a single path from left to right (e.g., event \(A\) AND event \(B\)), you multiply the probabilities.
2. Add DOWN paths: When an event can happen in more than one way (path \(1\) OR path \(2\)), you calculate each path's probability and add them together.
Key Takeaway: "AND means Multiply across; OR means Add down!"
3. Core Probability Rules on Tree Diagrams
Rule 1: Multiplication Rule (Along Branches)
To find the probability of a combined sequence of events along a path:
• For independent events: \(P(A \text{ and } B) = P(A) \times P(B)\)
• For dependent events: \(P(A \text{ and } B) = P(A) \times P(B \mid A)\)
Rule 2: Addition Rule (Across Multiple Paths)
When an outcome can happen via multiple separate paths, find each path's probability and add them:
\(P(\text{Event}) = \sum P(\text{favourable paths})\)
Rule 3: Total Probability Check
If you have worked out every path correctly, the sum of all final path probabilities must equal \(1\):
\(\sum P(\text{all paths}) = 1\)
Tip: Use this at the end of an exam question as a quick sanity check to guarantee full marks!
Rule 4: The Complement Rule ("At Least One")
CCEA examiners love asking for the probability of getting "at least one" specific item. Instead of adding three or four separate paths, use the complement shortcut:
\(P(\text{at least one}) = 1 - P(\text{none})\)
Key Takeaway: Total probability is always \(1\). Use \(1 - P(\text{none})\) to save time when finding "at least one".
4. Independent Events: "With Replacement" (M6 / M7 / M8)
Let's look at a step-by-step example where items are replaced, meaning the total number of items never changes.
Worked Example 1: Independent Selection
A bag contains \(3\) red counters and \(7\) blue counters (total \(10\) counters). A counter is chosen at random, its colour is noted, and it is replaced. A second counter is then chosen.
Step 1: Set up the branches
• 1st Pick:
Branch Red (\(R\)): \(P(R) = \frac{3}{10}\)
Branch Blue (\(B\)): \(P(B) = \frac{7}{10}\)
Check: \(\frac{3}{10} + \frac{7}{10} = 1\)
• 2nd Pick (Probabilities stay identical because the counter was replaced):
From 1st Red \(\rightarrow\) Red: \(\frac{3}{10}\), Blue: \(\frac{7}{10}\)
From 1st Blue \(\rightarrow\) Red: \(\frac{3}{10}\), Blue: \(\frac{7}{10}\)
Step 2: List all end outcomes and multiply along each path
• Path 1: \(P(RR) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}\)
• Path 2: \(P(RB) = \frac{3}{10} \times \frac{7}{10} = \frac{21}{100}\)
• Path 3: \(P(GR \text{ / } BR) = \frac{7}{10} \times \frac{3}{10} = \frac{21}{100}\)
• Path 4: \(P(BB) = \frac{7}{10} \times \frac{7}{10} = \frac{49}{100}\)
Check total sum: \(\frac{9}{100} + \frac{21}{100} + \frac{21}{100} + \frac{49}{100} = \frac{100}{100} = 1\)
Step 3: Answer specific exam questions
• Question A: Find the probability of picking two red counters.
\(P(RR) = \frac{9}{100}\)
• Question B: Find the probability of picking one counter of each colour.
"One of each" means we could pick Red then Blue (\(RB\)) OR Blue then Red (\(BR\)). We add both paths:
\(P(\text{one of each}) = P(RB) + P(BR) = \frac{21}{100} + \frac{21}{100} = \frac{42}{100} = \frac{21}{50}\)
Key Takeaway: With replacement, denominators and numerators remain constant on the second set of branches.
5. Dependent Events: "Without Replacement" (M8 Focus)
In conditional probability, the first item is not returned to the bag. This creates two important changes on the second set of branches:
1. The denominator decreases by \(1\) (from \(n\) to \(n - 1\)) because there is one fewer item in total.
2. The numerator decreases by \(1\) for whichever colour was just taken, while the numerator for the other colour stays the same (but sits over the new denominator \(n - 1\)).
Worked Example 2: Dependent Selection (Conditional)
A box contains \(4\) milk chocolates and \(6\) dark chocolates (total \(10\) chocolates). Sarah eats one chocolate at random and then eats a second chocolate.
Step 1: Set up the 1st Stage Branches
• Milk (\(M\)): \(P(M) = \frac{4}{10}\)
• Dark (\(D\)): \(P(D) = \frac{6}{10}\)
Step 2: Set up the 2nd Stage Branches (Total remaining = \(9\))
• If Sarah picked Milk first:
There are now \(3\) milk chocolates left and \(6\) dark chocolates left out of \(9\).
Branch Milk: \(P(M \mid M) = \frac{3}{9}\)
Branch Dark: \(P(D \mid M) = \frac{6}{9}\)
Check: \(\frac{3}{9} + \frac{6}{9} = 1\)
• If Sarah picked Dark first:
There are now \(4\) milk chocolates left and \(5\) dark chocolates left out of \(9\).
Branch Milk: \(P(M \mid D) = \frac{4}{9}\)
Branch Dark: \(P(D \mid D) = \frac{5}{9}\)
Check: \(\frac{4}{9} + \frac{5}{9} = 1\)
Step 3: Calculate Path Probabilities
• Path 1 (\(MM\)): \(\frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\)
• Path 2 (\(MD\)): \(\frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\)
• Path 3 (\(DM\)): \(\frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\)
• Path 4 (\(DD\)): \(\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\)
Step 4: Solve Typical Exam Questions
• Question A: What is the probability that Sarah eats two chocolates of the same colour?
"Same colour" means both Milk (\(MM\)) OR both Dark (\(DD\)):
\(P(\text{same colour}) = P(MM) + P(DD) = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}\)
• Question B: What is the probability that Sarah eats at least one dark chocolate?
Using our complement shortcut:
\(P(\text{at least one Dark}) = 1 - P(\text{no Dark}) = 1 - P(MM)\)
\(P(\text{at least one Dark}) = 1 - \frac{12}{90} = \frac{78}{90} = \frac{13}{15}\)
Key Takeaway: For "without replacement", always drop the denominator by \(1\) for the second choice, and reduce the numerator for whichever item was taken first!
6. Pitfalls & Common Examiner-Reported Errors
Make sure you avoid these common traps highlighted in CCEA Chief Examiner reports:
• Trap 1: Adding along branches instead of multiplying.
Incorrect: \(P(A \text{ and } B) = \frac{3}{10} + \frac{2}{9}\)
Correct: \(P(A \text{ and } B) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90}\)
• Trap 2: Forgetting to drop the denominator for dependent events.
Always check if the question says "without replacement", "eats", "takes two together", or "does not put it back". If so, your second denominator must be \(n - 1\).
• Trap 3: Missing the second order/combination for "one of each".
If asked for "one red and one green", calculating only \(P(RG)\) gives you only half the marks! You must include both \(P(RG)\) and \(P(GR)\) and add them together.
• Trap 4: Branches at a single node not adding to \(1\).
Always double-check each fork: \(\frac{3}{7} + \frac{4}{7} = 1\). If they don't add to \(1\), re-check your subtraction.
• Trap 5: Premature rounding of decimals.
Leaving answers as exact fractions (e.g. \(\frac{42}{90}\) or simplified \(\frac{7}{15}\)) avoids rounding errors and guarantees maximum accuracy on both Paper 1 (Non-Calculator) and Paper 2 (Calculator).
7. Quick Review Summary
• Branches meeting at any node: Sum must equal \(1\).
• Moving along a path (AND): Multiply the branch probabilities.
• Combining different paths (OR): Add the path outcomes.
• With replacement: Probabilities stay the same on stage 2.
• Without replacement: Total denominator decreases by \(1\); numerator of the chosen item decreases by \(1\).
• At least one: Calculate \(1 - P(\text{none})\).
• All final paths combined: Must sum to exactly \(1\).