Independent Events
Every question with a fully worked solution.
- Independent Events - Video - Probability of Independent and Dependent Events Watch
Theory
Two events are independent if one does not change the probability of the other, and dependent if it does. This Year 10 maths page shows how drawing with or without replacement decides which you have, how to use tree diagrams and tables, multiply along paths and use the complement for "at least one", and how to test two events for independence with \(P(A\cap B)=P(A)\times P(B)\).
A multistage experiment has two or more steps, such as drawing two counters from a bag or tossing a coin three times.
Two events are independent if the result of the first does not change the probability of the second. Coin tosses, dice rolls and draws with replacement (the item is put back) are independent.
Two events are dependent if the first result changes the probability of the second. Draws without replacement are dependent: after the first draw there is one item fewer, and one fewer of that colour. Choosing people for different jobs, eating lollies and dealing cards are all without replacement.
To find the probability of a sequence of results, multiply along the path. When an event can happen in more than one way, find each path and add them.
The complement of an event is everything else that can happen: \(P(\text{not } A) = 1 - P(A)\). It is the quick way to find "at least one".
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 events:
For dependent events, use the probability of \(B\) after \(A\) has happened:
The complement:
| With replacement | Without replacement | |
|---|---|---|
| Events are | independent | dependent |
| Second draw total | same as the first | one fewer |
| Same item twice? | possible | impossible |
| Bag of \(n\), two draws | \(n \times n\) outcomes | \(n \times (n-1)\) outcomes |
How to solve a multistage probability problem
- Decide: with replacement (independent) or without (dependent)? Words like "eaten", "chosen for a different job" and "not put back" mean without.
- Draw a tree diagram or a table of outcomes. Without replacement, take one off the total and one off the colour already drawn.
- Multiply along each path you want.
- Add the paths if the event can happen in more than one way.
- Simplify the fraction and check it is between \(0\) and \(1\).
Shortcuts
- At least one: find \(P(\text{none})\) and subtract it from \(1\).
- Only the second draw: if you are told the first result, just recount the bag. After a blue is removed from \(3\) red and \(4\) blue, \(P(\text{red}) = \dfrac{3}{6}\).
- Numbers from cards: list every outcome in a table, then count the ones you want.
(i) \(P(\text{both green})\)
(ii) \(P(\text{green, then yellow})\)
The token goes back, so the bag is the same for both draws. The draws are independent: multiply.
| \(P(GG)\) | \(=\) | \(\dfrac{4}{10} \times \dfrac{4}{10}\) |
| \(\) | \(=\) | \(\dfrac{16}{100} = \mathbf{\dfrac{4}{25}}\) |
| \(P(GY)\) | \(=\) | \(\dfrac{4}{10} \times \dfrac{6}{10}\) |
| \(\) | \(=\) | \(\dfrac{24}{100} = \mathbf{\dfrac{6}{25}}\) |
"Green, then yellow" fixes the order, so there is only one path.
(i) \(P(\text{both milk})\)
(ii) \(P(\text{one of each kind})\)
An eaten chocolate is not replaced, so the second draw has \(9\) left. Draw the tree and change the second branches.
(i) Multiply along the path MM.
| \(P(MM)\) | \(=\) | \(\dfrac{4}{10} \times \dfrac{3}{9}\) |
| \(\) | \(=\) | \(\dfrac{12}{90} = \mathbf{\dfrac{2}{15}}\) |
(ii) One of each is two paths, DM or MD. Multiply along each, then add.
| \(P(DM)\) | \(=\) | \(\dfrac{6}{10} \times \dfrac{4}{9}=\dfrac{24}{90}\) |
| \(P(MD)\) | \(=\) | \(\dfrac{4}{10} \times \dfrac{6}{9}=\dfrac{24}{90}\) |
| \(P\) | \(=\) | \(\dfrac{24}{90} + \dfrac{24}{90}\) |
| \(\) | \(=\) | \(\dfrac{48}{90} = \mathbf{\dfrac{8}{15}}\) |
(i) on exactly one day
(ii) on at least one day.
The days are independent. Let L = late and O = on time, so \(P(\text{O}) = 1 - 0.2 = 0.8\).
(i) Exactly one late day can happen in three orders: LOO, OLO and OOL. Each order has the same probability.
| \(P(\text{LOO})\) | \(=\) | \(0.2 \times 0.8 \times 0.8 = 0.128\) |
| \(P(\text{exactly one})\) | \(=\) | \(3 \times 0.128\) |
| \(\) | \(=\) | \(\mathbf{0.384}\) |
(ii) "At least one" is every outcome except "none". Use the complement.
| \(P(\text{none late})\) | \(=\) | \(0.8 \times 0.8 \times 0.8 = 0.512\) |
| \(P(\text{at least one})\) | \(=\) | \(1 - 0.512\) |
| \(\) | \(=\) | \(\mathbf{0.488}\) |
(i) greater than \(25\)
(ii) even.
Without replacement a digit cannot repeat, so the diagonal of the table is empty. There are \(4 \times 3 = 12\) outcomes.
| Tens / Units | 1 | 2 | 5 | 8 |
|---|---|---|---|---|
| 1 | — | 12 | 15 | 18 |
| 2 | 21 | — | 25 | 28 |
| 5 | 51 | 52 | — | 58 |
| 8 | 81 | 82 | 85 | — |
(i) Greater than \(25\) (bold): \(28, 51, 52, 58, 81, 82, 85\).
| \(P(>25)\) | \(=\) | \(\mathbf{\dfrac{7}{12}}\) |
(ii) Even means the units digit is \(2\) or \(8\): \(12, 18, 52, 58, 82, 28\).
| \(P(\text{even})\) | \(=\) | \(\dfrac{6}{12} = \mathbf{\dfrac{1}{2}}\) |
With replacement there would be \(4 \times 4 = 16\) outcomes, including \(11, 22, 55\) and \(88\).
Common pitfalls
Frequently asked questions
What is the difference between dependent and independent events?
Independent events do not affect each other, such as two coin tosses or draws with replacement. Dependent events do, such as draws without replacement, where the first draw changes what is left for the second.
How do you find the probability of two independent events?
Multiply their probabilities: \(P(A \text{ and } B) = P(A) \times P(B)\). For two heads, \(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
Is drawing without replacement dependent or independent?
Dependent. Once an item is removed, the total and the number of that type both drop, so the probabilities for the next draw change.
How do you work out at least one in probability?
Find the probability of none, then subtract it from \(1\). With a \(0.2\) chance on each of three independent days, \(1 - 0.8^3 = 0.488\).
When do you add probabilities on a tree diagram?
Add when the event can happen along more than one path, such as one of each colour (red then blue, or blue then red). Multiply along each path first, then add the results.
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 \(0\).