This note covers probabilities of combined events: the addition rule for "A or B", mutually exclusive events, conditional probability, and independent events. You will use Venn diagrams, tree diagrams and formulae to solve problems with and without replacement.
Key ideas
Combined events: A or B. To find the probability that \(A\) or \(B\) (or both) happens, add the probabilities and subtract the overlap, which would otherwise be counted twice: \[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \]
Mutually exclusive events cannot happen at the same time, so \(P(A\cap B)=0\). The rule then simplifies to \(P(A\cup B)=P(A)+P(B)\). For example, rolling a 2 and rolling a 5 on one die are mutually exclusive.
Conditional probability. \(P(A\mid B)\) means "the probability of \(A\), given that \(B\) has happened". Knowing \(B\) happened shrinks the sample space to \(B\), so \[ P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)\gt0. \] Rearranged, \(P(A\cap B)=P(B)\times P(A\mid B)\). On a tree diagram this is the rule "multiply along the branches": the second-stage branches carry conditional probabilities.
Independent events. Two events are independent if one happening does not change the probability of the other. Then \[ P(A\cap B)=P(A)\times P(B),\qquad\text{and equivalently}\qquad P(A\mid B)=P(A)=P(A\mid B'). \] To test for independence, check whether \(P(A\cap B)\) equals \(P(A)\times P(B)\), or whether \(P(A\mid B)\) equals \(P(A)\).
Mutually exclusive is not the same as independent. If \(A\) and \(B\) are mutually exclusive and both can happen on their own, then knowing \(B\) occurred tells you that \(A\) did not. So they are dependent.
Conditional probability from a Venn diagram. When the numbers of items are in a Venn diagram, \(P(A\mid B)\) is the number in \(A\cap B\) divided by the total number in \(B\). You are only looking inside circle \(B\), so everything outside it is ignored. For example, if 8 of the 20 items in \(B\) are also in \(A\), then \(P(A\mid B)=\tfrac{8}{20}=0.4\).
Reading the notation. The vertical bar means "given that". \(P(A\mid B)\) and \(P(B\mid A)\) are usually different. For instance, the probability that a person who plays football is a student is not the same as the probability that a student plays football. Check which event is given, because that event goes in the denominator.
Using the complement with combined events. "At least one" of several independent events is often easiest as \(1-P(\text{none})\). For two independent events, \(P(\text{none})=P(A')\times P(B')\). For example, if two independent machines fail with probabilities 0.1 and 0.2, then \(P(\text{at least one fails})=1-0.9\times0.8=0.28\).
Tree diagrams.
- Multiply the probabilities along a path to get the probability of that combination.
- Add the probabilities of different paths when the event can happen in more than one way.
- With replacement, the second-stage probabilities are the same as the first. Without replacement, the numbers in the second stage change because the total and the composition have changed.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) | Probability of \(A\) or \(B\) (or both). |
| Mutually exclusive: \(P(A\cap B)=0\) | The events cannot both happen. Then \(P(A\cup B)=P(A)+P(B)\). |
| \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) | Probability of \(A\) given \(B\) has happened. |
| \(P(A\cap B)=P(B)\,P(A\mid B)\) | Multiplying along a tree diagram. |
| Independent: \(P(A\cap B)=P(A)P(B)\) | The events do not affect each other. Equivalent to \(P(A\mid B)=P(A)\). |
Worked examples
Example 1 For events \(A\) and \(B\), \(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.2\). (a) Find \(P(A\cup B)\). (b) Find \(P(A\mid B)\). (c) Are \(A\) and \(B\) independent?
(a) \(P(A\cup B)=0.5+0.4-0.2=0.7\).
(b) \(P(A\mid B)=\dfrac{0.2}{0.4}=0.5\).
(c) \(P(A)\times P(B)=0.5\times0.4=0.2=P(A\cap B)\). Equivalently \(P(A\mid B)=0.5=P(A)\). So the events are independent.
Answer: (a) 0.7; (b) 0.5; (c) yes, independent
Example 2 A bag has 6 red and 4 blue counters. Two counters are drawn at random without replacement. Find (a) \(P(\text{both red})\), (b) \(P(\text{one of each colour})\), (c) \(P(\text{first is red}\mid\text{second is red})\).
Draw a tree. First draw: R with \(\tfrac6{10}\), B with \(\tfrac4{10}\). After a red, the bag has 5 red and 4 blue (9 counters). After a blue, it has 6 red and 3 blue.
(a) \(P(RR)=\dfrac{6}{10}\times\dfrac59=\dfrac{30}{90}=\dfrac13\).
(b) \(P(RB)=\dfrac6{10}\times\dfrac49=\dfrac{24}{90}\) and \(P(BR)=\dfrac4{10}\times\dfrac69=\dfrac{24}{90}\). Add the two paths: \(\dfrac{48}{90}=\dfrac{8}{15}\).
(c) The second counter is red in the paths RR and BR: \(P(\text{second red})=\dfrac{30}{90}+\dfrac{24}{90}=\dfrac{54}{90}=\dfrac35\).
\(P(\text{first red}\mid\text{second red})=\dfrac{P(RR)}{P(\text{second red})}=\dfrac{30/90}{54/90}=\dfrac{30}{54}=\dfrac59\).
Answer: (a) \(\tfrac13\); (b) \(\tfrac{8}{15}\); (c) \(\tfrac59\)
Example 3 Events \(A\) and \(B\) have \(P(A)=0.3\) and \(P(B)=0.5\). (a) Find \(P(A\cup B)\) if they are mutually exclusive. (b) Find \(P(A\cup B)\) if they are independent. (c) Can they be both?
(a) Mutually exclusive: \(P(A\cap B)=0\), so \(P(A\cup B)=0.3+0.5=0.8\).
(b) Independent: \(P(A\cap B)=0.3\times0.5=0.15\). So \(P(A\cup B)=0.3+0.5-0.15=0.65\).
(c) Mutually exclusive needs \(P(A\cap B)=0\). Independent needs \(P(A\cap B)=0.15\). These cannot both be true.
Answer: (a) 0.8; (b) 0.65; (c) no
Common errors
- Common error: forgetting to subtract \(P(A\cap B)\) when the events can overlap, so the union probability is too large.
- Common error: confusing "mutually exclusive" with "independent". Exclusive events with non-zero probabilities are never independent.
- Common error: dividing by the wrong probability in a conditional probability. In \(P(A\mid B)\), divide by \(P(B)\), the event that is given.
- Common error: not changing the second-stage fractions on a tree diagram when there is no replacement.
- Common error: multiplying where you should add. Multiply along a path; add between different paths.
Exam tip: To show that events are independent, write down both \(P(A)\times P(B)\) and \(P(A\cap B)\) with their values and state that they are equal. A statement without numbers is not a proof.
Check yourself
- \(P(A)=0.6\), \(P(B)=0.35\), \(P(A\cap B)=0.15\). Find \(P(A\cup B)\).
Answer
\(0.6+0.35-0.15=0.8\). - \(P(A\cap B)=0.12\) and \(P(B)=0.4\). Find \(P(A\mid B)\).
Answer
\(\dfrac{0.12}{0.4}=0.3\). - \(X\) and \(Y\) are independent with \(P(X)=0.7\) and \(P(Y)=0.2\). Find \(P(X\cap Y)\) and \(P(X\cup Y)\).
Answer
\(P(X\cap Y)=0.7\times0.2=0.14\). \(P(X\cup Y)=0.7+0.2-0.14=0.76\). - \(P(A)=0.5\), \(P(B)=0.3\) and \(P(A\cup B)=0.65\). Are \(A\) and \(B\) independent?
Answer
\(P(A\cap B)=0.5+0.3-0.65=0.15\). \(P(A)P(B)=0.5\times0.3=0.15\). These are equal, so the events are independent.