Dependent and Independent Events
Two events are independent if one happening does not change the probability of the other, so \(P(A\text{ and }B)=P(A)\times P(B)\). Events are dependent when it does, as when a card is drawn and not replaced. In NSW Year 10 Advanced maths you decide whether events are independent and calculate probabilities for both kinds (NESA outcome MA5‑PRO‑C‑01).
You will compare drawing with and without replacement, test independence using the multiplication rule with dice, cards and survey data, and find the probability that at least one or exactly one of two independent events happens. You will choose a captain, vice-captain and secretary from a class, win a voucher in a lucky dip, and find an unknown probability from \(P(A\text{ or }B)\). One bag gains an extra marble after each draw. It is part of Year 10 Advanced Probability.
Every question with a fully worked solution.
- Dependent and Independent Events - Video - Probability of Independent and Dependent Events Watch
Theory
Two events are independent if one happening does not change the probability of the other. Then \(P(A\cap B)=P(A)\times P(B)\). This Year 10 Advanced maths page covers testing for independence with numbers, independence from survey counts, calculating with independent events, dependent draws without replacement, and the difference from mutually exclusive events.
Events \(A\) and \(B\) are independent if the occurrence of one does not affect the probability of the other: \(P(A\mid B)=P(A)\). Two coin tosses, or two draws with replacement, are independent.
Events are dependent if one changes the probability of the other. Draws without replacement are dependent, because the first draw changes what is left.
The multiplication rule for independent events is \(P(A\cap B)=P(A)\times P(B)\). It is also the test: if the product equals \(P(A\cap B)\), the events are independent; if not, they are dependent. Events from one experiment can be independent too, such as "less than \(3\)" and "odd" on a single die.
Independent is not the same as mutually exclusive. Mutually exclusive events cannot happen together, so \(P(A\cap B)=0\). If both probabilities are positive, \(P(A)\times P(B)\gt 0\), so mutually exclusive events are always dependent.
| For independent \(A\), \(B\) | Formula |
|---|---|
| both | \(P(A)P(B)\) |
| at least one | \(P(A)+P(B)-P(A)P(B)\) \(=1-P(A')P(B')\) |
| neither | \(P(A')P(B')\) \(=(1-P(A))(1-P(B))\) |
| exactly one | \(P(A)P(B')+P(A')P(B)\) |
How to decide and calculate
- Find \(P(A)\), \(P(B)\) and \(P(A\cap B)\), from a list, a table, counts or the addition rule.
- Compare \(P(A)\times P(B)\) with \(P(A\cap B)\). Equal means independent; different means dependent. State the comparison as your reason.
- If independent, multiply probabilities for "and" and use complements for "at least one".
- If dependent, use a tree with conditional second-stage branches.
Other question types
- Find an unknown for independence: set \(P(A\cap B)=P(A)P(B)\). With a table, \(\dfrac{x}{\text{total}}=\) (row proportion) \(\times\) (column proportion).
- Find a missing probability: write \(P(A\cup B)=P(A)+p-P(A)p\) and solve (Example 3). With \(P(B)\) written in terms of \(p\), this can become a quadratic; reject solutions that make a probability greater than \(1\).
- Compare the two cases: for the same \(P(A)\) and \(P(B)\), mutually exclusive gives \(P(A\cup B)=P(A)+P(B)\); independent gives \(P(A)+P(B)-P(A)P(B)\).
- Two people's results (such as two friends both late) are dependent if the question gives a different probability for the second person depending on the first.
(i) A fair die is rolled. \(A\) is "less than \(3\)" and \(B\) is "odd". Are \(A\) and \(B\) independent?
(ii) A card is drawn from a standard pack. \(H\) is "a heart" and \(R\) is "a red card". Are \(H\) and \(R\) independent?
(i) \(A=\{1, 2\}\), \(B=\{1, 3, 5\}\), \(A\cap B=\{1\}\).
| \(P(A)\times P(B)\) | \(=\) | \(\dfrac{2}{6}\times\dfrac{3}{6}=\dfrac{1}{6}\) |
| \(P(A\cap B)\) | \(=\) | \(\dfrac{1}{6}\) |
They are equal, so \(A\) and \(B\) are independent.
(ii) Every heart is red, so \(P(H\cap R)=P(H)=\dfrac{1}{4}\).
| \(P(H)\times P(R)\) | \(=\) | \(\dfrac{1}{4}\times\dfrac{1}{2}=\dfrac{1}{8}\neq\dfrac{1}{4}\) |
So \(H\) and \(R\) are dependent: knowing the card is red doubles the chance it is a heart.
(i) Are "has a pet" and "learns an instrument" independent?
(ii) How many would need to do both for the events to be independent?
(i) Convert the counts to probabilities.
| \(P(\text{pet})\) | \(=\) | \(\dfrac{180}{300}=0.6\) |
| \(P(\text{instrument})\) | \(=\) | \(\dfrac{90}{300}=0.3\) |
| \(P(\text{both})\) | \(=\) | \(\dfrac{60}{300}=0.2\) |
\(0.6\times 0.3=0.18\neq 0.2\), so the events are dependent.
(ii) Independence needs \(P(\text{both})=0.18\), which is \(0.18\times 300=\boldsymbol{54}\) students.
(i) \(A\) and \(B\) are independent, with \(P(A)=0.6\) and \(P(B)=0.35\). Find \(P(A\cap B)\), \(P(A\cup B)\), \(P(\text{exactly one})\) and \(P(\text{neither})\).
(ii) \(C\) and \(D\) are independent, with \(P(C)=0.4\) and \(P(C\cup D)=0.7\). Find \(P(D)\).
(i) Multiply for "and", then use the addition rule.
| \(P(A\cap B)\) | \(=\) | \(0.6\times 0.35=\boldsymbol{0.21}\) |
| \(P(A\cup B)\) | \(=\) | \(0.6+0.35-0.21=\boldsymbol{0.74}\) |
| \(P(\text{exactly one})\) | \(=\) | \(0.74-0.21=\boldsymbol{0.53}\) |
| \(P(\text{neither})\) | \(=\) | \(0.4\times 0.65=\boldsymbol{0.26}\) |
(ii) Let \(P(D)=p\). Independence gives \(P(C\cap D)=0.4p\).
| \(0.4+p-0.4p\) | \(=\) | \(0.7\) |
| \(0.6p\) | \(=\) | \(0.3\) |
| \(p\) | \(=\) | \(\boldsymbol{0.5}\) |
(i) the probability that the second student chosen is a boy
(ii) \(P(\text{exactly one boy})\)
(i) The second choice depends on the first. Add the two paths ending in a boy (Diagrams tab).
| \(P\) | \(=\) | \(\dfrac{8}{15}\times\dfrac{7}{14}+\dfrac{7}{15}\times\dfrac{6}{14}\) |
| \(\) | \(=\) | \(\dfrac{56+42}{210}=\boldsymbol{\dfrac{7}{15}}\) |
(ii) Exactly one boy: \(BGG\), \(GBG\) or \(GGB\). Each has the same three numerators in a different order.
| \(P(BGG)\) | \(=\) | \(\dfrac{7}{15}\times\dfrac{8}{14}\times\dfrac{7}{13}=\dfrac{392}{2730}\) |
| \(P\) | \(=\) | \(3\times\dfrac{392}{2730}=\boldsymbol{\dfrac{28}{65}}\) |
Common pitfalls
Frequently asked questions
What is the difference between dependent and independent events?
Independent events do not affect each other's probabilities, such as two coin tosses. Dependent events do, such as drawing two cards without replacement, where the first card changes what is left.
How do you test whether two events are independent?
Work out \(P(A)\times P(B)\) and compare it with \(P(A\cap B)\). If they are equal, the events are independent. You can also check whether \(P(A\mid B)=P(A)\).
Can events be both independent and mutually exclusive?
Not if both have a positive probability. Mutually exclusive means \(P(A\cap B)=0\), but independence needs \(P(A\cap B)=P(A)P(B)\), which is more than \(0\).
Are draws with replacement independent?
Yes. Putting the item back restores the original contents, so every draw has the same probabilities whatever happened before.
How do you find the probability of at least one of two independent events?
Subtract the probability that neither happens from \(1\). Neither is \((1-P(A))(1-P(B))\) because the events are independent.