This note shows how to find the probability of two or more events happening together. You will use the multiplication and addition rules, draw tree diagrams for events with and without replacement, and calculate conditional probabilities from trees, Venn diagrams and tables. These methods are used in medical testing, risk assessment and quality control.
Key ideas
Independent events. Two events are independent if one happening does not change the chance of the other. For example, two separate coin tosses are independent. For independent events:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
Mutually exclusive events. Two events that cannot happen together are mutually exclusive. For these:
\[ P(A \text{ or } B) = P(A) + P(B) \]
The rule of thumb is: "and" means multiply along a branch, "or" means add the branches.
Tree diagrams. A tree diagram shows each stage of an experiment as a set of branches. Write the probability on each branch.
- The probabilities on the branches leaving any one point add up to 1.
- To find the probability of a path, multiply the probabilities along its branches.
- To find the probability of an event that can happen in several ways, add the probabilities of all the paths that give it.
- The probabilities of all the complete paths add up to 1, which is a good check.
With replacement and without replacement. If an item is put back after being picked (with replacement), the probabilities on the second set of branches are the same as on the first, and the events are independent. If it is not put back (without replacement), the number of items goes down by one, so the second-stage probabilities change. The events are then dependent.
Conditional probability. The probability of event \(B\) given that event \(A\) has already happened is written \(P(B \mid A)\). In a tree diagram, the branches of the second stage are already conditional probabilities. For other situations, restrict your attention to the cases where \(A\) happened, and count within them:
\[ P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)} \]
With a Venn diagram or a table of numbers, this means: the numerator is the number in both \(A\) and \(B\), and the denominator is the total number in \(A\). Notice that \(P(B \mid A)\) and \(P(A \mid B)\) are usually different, because they use different totals.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(P(A \text{ and } B) = P(A) \times P(B)\) | Independent events happening together. |
| \(P(A \text{ or } B) = P(A) + P(B)\) | Mutually exclusive events: either one happening. |
| Independent events | The outcome of one does not affect the other. |
| Dependent events | The outcome of one changes the probability of the other (for example, without replacement). |
| \(P(B \mid A)\) | Probability of \(B\) given that \(A\) has happened. |
| \(P(B \mid A) = \dfrac{P(A \text{ and } B)}{P(A)}\) | Conditional probability from a tree, Venn diagram or table. |
Worked examples
Example 1 The probability of rain on any day in a town is 0.3, independently from day to day. Find the probability that over two days (a) it rains on both days, (b) it rains on exactly one day.
Draw a tree with two stages. Each stage has a branch for rain (0.3) and a branch for no rain (0.7).
(a) \(P(\text{rain, rain}) = 0.3 \times 0.3 = 0.09\).
(b) There are two paths with exactly one rainy day: rain then dry, and dry then rain.
\(P(\text{rain, dry}) = 0.3 \times 0.7 = 0.21\) and \(P(\text{dry, rain}) = 0.7 \times 0.3 = 0.21\).
Add the paths: \(0.21 + 0.21 = 0.42\).
Answer: (a) 0.09. (b) 0.42.
Example 2 A bag contains 5 red and 3 blue counters. Two counters are taken out one after the other without replacement. Find (a) the probability that both are the same colour, and (b) the probability that the first counter is red given that the second is blue.
First stage: \(P(R) = \tfrac58\) and \(P(B) = \tfrac38\).
Second stage after a red: 4 red and 3 blue remain out of 7, so \(P(R) = \tfrac47\) and \(P(B) = \tfrac37\).
Second stage after a blue: 5 red and 2 blue remain out of 7, so \(P(R) = \tfrac57\) and \(P(B) = \tfrac27\).
(a) Both red: \(\tfrac58 \times \tfrac47 = \tfrac{20}{56}\). Both blue: \(\tfrac38 \times \tfrac27 = \tfrac{6}{56}\).
Same colour: \(\tfrac{20}{56} + \tfrac{6}{56} = \tfrac{26}{56} = \tfrac{13}{28}\).
(b) Red then blue: \(\tfrac58 \times \tfrac37 = \tfrac{15}{56}\). Blue then blue: \(\tfrac{6}{56}\).
\(P(\text{second is blue}) = \tfrac{15}{56} + \tfrac{6}{56} = \tfrac{21}{56}\).
\(P(\text{first red} \mid \text{second blue}) = \dfrac{15/56}{21/56} = \dfrac{15}{21} = \dfrac57\).
Answer: (a) \(\tfrac{13}{28}\). (b) \(\tfrac57\).
Example 3 In a class of 30 students, 18 play football, 12 play tennis and 5 play both. A student is chosen at random. (a) How many play neither sport? (b) Given that the student plays football, find the probability that they also play tennis. (c) Given that the student plays tennis, find the probability that they also play football.
Draw a Venn diagram. The overlap has 5. Football only: \(18 - 5 = 13\). Tennis only: \(12 - 5 = 7\).
(a) Students playing at least one sport: \(13 + 5 + 7 = 25\). Neither: \(30 - 25 = 5\).
(b) Restrict to the 18 football players. Of these, 5 also play tennis. \(P(\text{tennis} \mid \text{football}) = \dfrac{5}{18}\).
(c) Restrict to the 12 tennis players. Of these, 5 also play football. \(P(\text{football} \mid \text{tennis}) = \dfrac{5}{12}\).
Answer: (a) 5 students. (b) \(\tfrac{5}{18}\). (c) \(\tfrac{5}{12}\).
Common errors
- Common error: using the same probabilities on the second set of branches when the item is not replaced. The totals must go down by one.
- Common error: adding along a branch or multiplying between different paths. Multiply along, add between.
- Common error: forgetting a path. For "exactly one", there are usually two paths, not one.
- Common error: using the whole group as the denominator for a conditional probability, instead of only the group that is given.
- Common error: confusing \(P(B \mid A)\) with \(P(A \mid B)\).
Exam tip: After finishing a tree diagram, add up the probabilities of all complete paths. If they do not total 1, you have an arithmetic error or a wrong branch.
Check yourself
- A fair coin is tossed and a fair die is rolled. Find the probability of a head and a six.
Answer
The events are independent: \(\tfrac12 \times \tfrac16 = \tfrac{1}{12}\). - A bag has 4 yellow and 6 green beads. Two are taken without replacement. Find the probability that both are green.
Answer
\(\tfrac{6}{10} \times \tfrac{5}{9} = \tfrac{30}{90} = \tfrac13\). - The probability that a bus is on time is 0.8 on each of three independent days. Find the probability that it is on time on all three days.
Answer
\(0.8 \times 0.8 \times 0.8 = 0.512\). - For the class in Example 3, which is greater: \(P(\text{plays tennis} \mid \text{plays football})\) or \(P(\text{plays football} \mid \text{plays tennis})\)?
Answer
The first is \(\tfrac{5}{18} = 0.278\) and the second is \(\tfrac{5}{12} = 0.417\). The second is greater, because the smaller group (12 tennis players) is the one being restricted to.