Ch 2 · 8.3 Probability of Combined Events · IGCSE Mathematics 0580
Topic 2 of 3 in Probability — IGCSE Mathematics 0580 — 18 lessons.
Introduction
Two dice, two counters from a bag, two traffic lights on the way to school: when events combine, you need a systematic way to list or calculate the outcomes. This lesson uses three diagrams: sample spaces, Venn diagrams and tree diagrams. (Syllabus section 8.3.)
Key ideas
For two dice, a 6 × 6 grid lists all 36 equally likely outcomes. P(total 7) = 6/36 = tfrac16.
🧑🏫 Teacher's explanation
Imagine you are organizing a party and want to know the chances of certain guests arriving together. If you pick two people from a group, the outcome depends on whether you put the first person back into the group before picking the second (with replacement) or if they stay out of the group (without replacement). Combined events are just different ways of looking at these 'what if' scenarios.
📖 Definitions
🔍 Worked example — Easy
A bag contains 3 red marbles and 2 blue marbles. Two marbles are drawn one by one with replacement. Find the probability that both marbles are red.
🔍 Worked example — Standard
A box contains 5 green pens and 3 black pens. Two pens are drawn one by one without replacement. Find the probability that both pens are black.
🔍 Worked example — Exam-style
In a Venn diagram, P(A) = 0.6, P(B) = 0.5, and P(A ∩ B) = 0.2. Find P(A ∪ B).
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples
A bag has 5 red and 3 blue balls. Two are taken without replacement. P(both red) = tfrac58 × tfrac47 = 20/56 = 5/14. P(at least one blue) = 1 - 5/14 = 9/14. P(one of each) = tfrac58 × tfrac37 + tfrac38 × tfrac57 = 30/56 = 15/28.
🔍 Worked example — Easy
A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. Calculate the probability that both balls are red.
🔍 Worked example — Standard
In a class of 30 students, 12 study French (F), 15 study Spanish (S), and 6 study both. Calculate P(F ∪ S).
🔍 Worked example — Exam-style
A fair coin is tossed three times. Calculate the probability of getting at least one tail.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) (4,6),(5,5),(6,4),(5,6),(6,5),(6,6): 6/36 = tfrac16. (2) 2 × 0.2 × 0.8 = 0.32. (3) 4/10 × tfrac39 + 6/10 × tfrac59 = 12 + 30/90 = 7/15. (4) n(A ∪ B) = 37, so 13/50.