Probability of combined events
Section: Probability | Syllabus: Cambridge IGCSE Mathematics (0580)
The AND and OR Rules for Combined Events
- Events are independent if the outcome of one does not affect the outcome of the other (e.g. tossing a coin and spinning a spinner)
- AND rule (independent events): multiply the probabilities P(A and B)=P(A)× P(B)
- OR rule (mutually exclusive events): add the probabilities P(A or B)=P(A)+P(B)
- These two rules are often combined in the same question - work out an "or" probability first if needed, then multiply it by an independent "and" probability
Worked Example: Combining the AND and OR Rules
- Question: A fair coin is tossed and a fair four-sided spinner numbered 1 to 4 is spun. Find the probability of getting a tail and a number greater than 2.
- Step 1 (OR rule): "Greater than 2" on the spinner means 3 or 4, which are mutually exclusive outcomes P(3 or 4)=(1/4)+(1/4)=(1/2)
- Step 2 (AND rule): The coin and the spinner are independent, so multiply P(tail and>2)=(1/2)×(1/2)=(1/4)
- Answer: P(tail and>2)=(1/4)
Common Mistakes
MistakeAdding probabilities for independent "and" events instead of multiplying
Fix"and" for independent events means multiply; "or" for mutually exclusive events means add - mixing these up gives an answer greater than 1 or otherwise impossible
MistakeUsing the OR rule (adding) for two events that are not actually mutually exclusive
Fixcheck the events cannot both happen at once before adding their probabilities directly
Tree Diagrams for Independent Events
- A tree diagram shows all possible outcomes of two or more events as branches, with the probability of each outcome written on its branch
- To find the probability of a particular path through the tree, multiply along the branches
- To find the probability of two or more different paths (e.g. "both the same" or "exactly one"), add the resulting path probabilities
- When an item is picked, its colour noted, and then replaced before the next pick, the probabilities stay the same on every branch of the tree
Worked Example: Using a Tree Diagram (With Replacement)
- Question: A bag contains 3 red and 2 blue counters. A counter is picked at random, its colour noted, and then replaced. A second counter is then picked. Find the probability that both counters are the same colour.
- Step 1: P(red)=(3/5) and P(blue)=(2/5) on every branch, since the counter is replaced
- Step 2: P(both red)=(3/5)×(3/5)=(9/25)
- Step 3: P(both blue)=(2/5)×(2/5)=(4/25)
- Step 4: "Both the same colour" means both red OR both blue, so add these two path probabilities P(same colour)=(9/25)+(4/25)=(13/25)
- Answer: P(same colour)=(13/25)
Common Mistakes
MistakeAdding along the branches of a single path instead of multiplying
Fixalways multiply along a single path through the tree; only add when combining two or more separate, complete paths
MistakeForgetting to include every path that satisfies the event before adding (e.g. missing "both blue" when finding "same colour")
Fixlist every complete path that matches the required outcome before adding - it is easy to only spot one of two or more qualifying paths
Combined Events Without Replacement
Extended Only
- When an item is picked and not replaced, the total number of items (and the number of a particular type) changes for the next pick, so the branch probabilities on a tree diagram change too
- For "exactly one of a certain type from two picks", there are usually two orders that satisfy this - both must be found and added together
Worked Example: Combined Events Without Replacement
- Question: A box contains 5 blue pens and 3 black pens. Two pens are taken at random without replacement. Find the probability that exactly one of the two pens is black.
- Step 1: "Exactly one black" can happen in two orders: black-then-blue, or blue-then-black
- Step 2 (black then blue): (3/8)×(5/7)=(15/56)
- Step 3 (blue then black): (5/8)×(3/7)=(15/56)
- Step 4: Add the two orders (15/56)+(15/56)=(30/56)=(15/28)
- Answer: P(exactly one black)=(15/28)
Common Mistakes
MistakeUsing the same denominator for the second pick as the first, as if the item had been replaced
Fixwithout replacement, the total number of items decreases by 1 for the second pick - and the count of whichever type was picked first also decreases by 1
MistakeOnly calculating one of the two orders for an "exactly one" question and forgetting to double or add the other order
Fix"exactly one" from two picks (without replacement) almost always needs both possible orders added together
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