This deep dive covers Step-by-Step Method within Tree Diagrams for GCSE Mathematics. Revise Tree Diagrams in Probability for GCSE Mathematics with 15 exam-style questions and 20 flashcards. This is a high-frequency topic, so it is worth revising until the explanation feels precise and repeatable. It is section 4 of 6 in this topic. Use this deep dive to connect the idea to the wider topic before moving on to questions and flashcards.
Topic position
Section 4 of 6
Practice
15 questions
Recall
20 flashcards
Step-by-Step Method
Step 1: Draw the First Stage
From a starting point, draw branches for each possible outcome of the first event. Label each branch with its probability.
Step 2: Draw Subsequent Stages
From the end of each branch, draw new branches for the next stage. Label with appropriate probabilities.
Step 3: Calculate Path Probabilities
Multiply probabilities along each complete path to find the probability of that specific sequence.
Step 4: Find Event Probabilities
Add the probabilities of all paths that lead to the event you're interested in.
Worked Example: Bag with Replacement
Scenario: A bag contains 3 red balls and 2 blue balls. Two balls are drawn with replacement.
Find: P(both balls are red)
Solution:
Step 1: P(red) = 3/5, P(blue) = 2/5
Step 2: Draw tree diagram
First Draw Second Draw Outcome Probability
R (3/5) ——— RR 3/5 × 3/5 = 9/25
R (3/5) ——————/
/ \
/ B (2/5) ——— RB 3/5 × 2/5 = 6/25
\
\
B (2/5) ——————\
/
/ R (3/5) ——— BR 2/5 × 3/5 = 6/25
\
B (2/5) ——— BB 2/5 × 2/5 = 4/25
Step 3: P(both red) = P(RR) = 9/25
Worked Example: Without Replacement
Scenario: Same bag, but two balls drawn without replacement.
Find: P(both balls are red)
Solution:
Step 1: First draw: P(red) = 3/5, P(blue) = 2/5
Step 2: Second draw depends on first:
- If first is red: 2 red, 2 blue remain → P(red) = 2/4, P(blue) = 2/4
- If first is blue: 3 red, 1 blue remain → P(red) = 3/4, P(blue) = 1/4
First Draw Second Draw Outcome Probability
R (2/4) ——— RR 3/5 × 2/4 = 6/20 = 3/10
R (3/5) ——————/
/ \
/ B (2/4) ——— RB 3/5 × 2/4 = 6/20 = 3/10
\
\
B (2/5) ——————\
/
/ R (3/4) ——— BR 2/5 × 3/4 = 6/20 = 3/10
\
B (1/4) ——— BB 2/5 × 1/4 = 2/20 = 1/10
Step 3: P(both red) = P(RR) = 3/10