Probability of Combined Events
Dependent events
Events where one happening changes the probability of the other.
| English | Dependent events |
|---|---|
| Bahasa Melayu | Peristiwa bersandar |
| 中文 | 相关事件 |
How it is used
A bag has 3 red and 5 blue balls; two are drawn without replacement. P(both red) = 3/8 × 2/7 = 6/56 = 3/28, because after taking one red only 2 reds remain out of 7 balls.
Where it shows up in SPM
The core of 'without replacement' problems, a favourite in Paper 2 structured questions. You draw a tree diagram in which the second set of branches uses updated fractions, then multiply along paths.
Marks are often lost by forgetting to reduce both the count and the total on the second draw.
Don't confuse it with
Open the chapter: Probability of Combined Events →
Frequently asked questions
How do I adjust the probabilities for the second draw?
Reduce the total by 1, and if the item drawn was the colour you are tracking, reduce that colour's count by 1 too. From 3 red out of 8, after drawing a red the next red probability is 2/7.
Redraw the second branches of your tree with these new fractions.
How can I tell if a question is dependent or independent?
Look for the phrase 'without replacement' or a context where the item is not returned, such as eating a sweet or giving out prizes; these are dependent. 'With replacement' or separate objects like two dice are independent.
If unsure, ask whether the second probability could change.
Do I still multiply along the branches for dependent events?
Yes. The multiplication rule still applies along each path; the only difference from independent events is that the second branch uses the updated fraction.
So P(red then blue) = 3/8 × 5/7 = 15/56, multiplying the first probability by the adjusted second one.