The Central Idea
Suppose \(A\) and \(B\) are events.
The event
\[ A\cap B \]means that both \(A\) and \(B\) occur.
The general multiplication rule is
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]Equivalently,
\[ \boxed{ P(A\cap B)=P(B)P(A\mid B) }. \]The first probability starts the path. The conditional probability describes what happens next after the first event is known.
Where the Rule Comes From
Conditional probability is defined by
\[ P(B\mid A)=\frac{P(A\cap B)}{P(A)} \]whenever \(P(A)>0\).
Multiply both sides by \(P(A)\):
\[ P(A)P(B\mid A)=P(A\cap B). \]Therefore,
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]The multiplication rule is thus a rearrangement of conditional probability.
The Rule Can Be Written in Either Direction
Starting from \(A\),
\[ P(A\cap B)=P(A)P(B\mid A). \]Starting from \(B\),
\[ P(A\cap B)=P(B)P(A\mid B). \]Since both expressions equal the same intersection probability,
\[ \boxed{ P(A)P(B\mid A) = P(B)P(A\mid B) }. \]The best form to use depends on which probabilities are known.
Intersection Means “And”
In probability language,
\[ P(A\cap B) \]is often read as
the probability of \(A\) and \(B\).
Words such as and, both, followed by, or then often signal that a multiplication rule may be relevant.
A First Example
Suppose
\[ P(A)=0.40 \]and
\[ P(B\mid A)=0.70. \]Then
\[ P(A\cap B) = P(A)P(B\mid A). \]Therefore,
\[ P(A\cap B) = 0.40(0.70) = \boxed{0.28}. \]Thus \(28\%\) of outcomes satisfy both events under this model.
The Independent-Events Multiplication Rule
If \(A\) and \(B\) are independent, then
\[ P(B\mid A)=P(B). \]Substitute this into the general multiplication rule:
\[ P(A\cap B)=P(A)P(B). \]Therefore, for independent events,
\[ \boxed{ P(A\cap B)=P(A)P(B) }. \]This is a special case of the general multiplication rule.
Independent Versus Dependent Forms
The general rule is
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]If the events are independent, then
\[ P(B\mid A)=P(B), \]so the rule simplifies to
\[ \boxed{ P(A\cap B)=P(A)P(B) }. \]For dependent events, the conditional probability must remain in the calculation.
Example: Independent Coin Tosses
A fair coin is tossed twice.
Let \(A\) be the event that the first toss is heads and \(B\) the event that the second toss is heads.
The tosses are independent, so
\[ P(A)=\frac{1}{2} \]and
\[ P(B)=\frac{1}{2}. \]Therefore,
\[ P(A\cap B) = \frac{1}{2}\cdot\frac{1}{2} = \boxed{\frac{1}{4}}. \]Example: Three Independent Coin Tosses
Suppose a fair coin is tossed three times. Find the probability of three heads.
Each toss has probability
\[ \frac{1}{2} \]of heads, and the tosses are independent.
Therefore,
\[ P(HHH) = \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \boxed{\frac{1}{8}}. \]Example: Repeated Independent Successes
Suppose a process succeeds with probability
\[ p=0.80 \]on each of three independent trials.
The probability that all three trials succeed is
\[ 0.80^3 = \boxed{0.512}. \]Repeated independent multiplication produces powers of a common probability.
Dependent Events Require Conditional Probabilities
When one event changes the probability of another, we cannot replace
\[ P(B\mid A) \]with
\[ P(B). \]We must use
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]Example: Two Cards Without Replacement
Two cards are drawn from a standard \(52\)-card deck without replacement. Find the probability that both cards are aces.
The probability that the first card is an ace is
\[ \frac{4}{52}. \]If the first card is an ace, \(3\) aces remain among \(51\) cards:
\[ P(\text{second ace}\mid\text{first ace}) = \frac{3}{51}. \]Therefore,
\[ P(\text{two aces}) = \frac{4}{52}\cdot\frac{3}{51}. \]Thus,
\[ \boxed{ P(\text{two aces})=\frac{1}{221} }. \]Why the Denominator Changes Without Replacement
After one card is drawn and not returned, only \(51\) cards remain.
If that card was an ace, only \(3\) aces remain.
Therefore the second-stage probability must reflect the updated state:
\[ \frac{3}{51}, \]not
\[ \frac{4}{52}. \]The changing denominator and numerator are consequences of dependence.
Example: Two Cards With Replacement
Now suppose the first card is returned to the deck before the second card is drawn.
Each draw then has
\[ P(\text{ace})=\frac{4}{52}=\frac{1}{13}. \]Because replacement restores the original deck, the draws are independent.
Therefore,
\[ P(\text{two aces}) = \frac{1}{13}\cdot\frac{1}{13} = \boxed{\frac{1}{169}}. \]Sequential Probability
Many multiplication problems describe a sequence:
\[ A_1,\ A_2,\ A_3,\ \ldots \]Each stage may depend on what happened before it.
For three events,
\[ \boxed{ P(A\cap B\cap C) = P(A)P(B\mid A)P(C\mid A\cap B) }. \]This is the three-event chain form of the multiplication rule.
The General Chain Rule
For events \(A_1,A_2,\ldots,A_n\),
\[ \boxed{ P(A_1\cap A_2\cap\cdots\cap A_n) = P(A_1) P(A_2\mid A_1) P(A_3\mid A_1\cap A_2) \cdots P(A_n\mid A_1\cap\cdots\cap A_{n-1}) }. \]Each factor represents the next stage conditioned on the path already taken.
Example: Three Draws Without Replacement
A bag contains \(5\) red balls and \(3\) blue balls. Three balls are drawn without replacement.
Find the probability that all three are red.
The first red has probability
\[ \frac{5}{8}. \]Given a red first draw, the second red has probability
\[ \frac{4}{7}. \]Given two red draws, the third red has probability
\[ \frac{3}{6}. \]Therefore,
\[ P(RRR) = \frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}. \]Simplifying,
\[ \boxed{ P(RRR)=\frac{5}{28} }. \]Example: A Specific Color Sequence
Using the same bag, find the probability of drawing red, then blue, then red without replacement.
First red:
\[ \frac{5}{8}. \]After a red is removed, \(3\) blue balls remain among \(7\):
\[ \frac{3}{7}. \]After red then blue, \(4\) red balls remain among \(6\):
\[ \frac{4}{6}. \]Therefore,
\[ P(RBR) = \frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6} = \boxed{\frac{5}{28}}. \]Multiplication Along a Probability Tree
A probability tree represents possible sequences as branches.
The probability of a complete path is found by multiplying the branch probabilities along that path.
For example, if
\[ P(A)=0.60 \]and
\[ P(B\mid A)=0.75, \]then the path \(A\) followed by \(B\) has probability
\[ 0.60(0.75)=\boxed{0.45}. \]Branch Probabilities Depend on the Node
In a probability tree, a second-stage branch is usually conditional on the branch that led to its current node.
For example,
\[ P(B\mid A) \]and
\[ P(B\mid A^c) \]may be different.
The tree should record the probability appropriate to each path, not one universal second-stage probability unless the model justifies it.
Example: Customer Conversion Tree
Suppose \(60\%\) of visitors reach a product page:
\[ P(P)=0.60. \]Of those who reach the page, \(25\%\) purchase:
\[ P(B\mid P)=0.25. \]The probability that a visitor reaches the page and purchases is
\[ P(P\cap B) = 0.60(0.25) = \boxed{0.15}. \]Thus \(15\%\) of all visitors follow this complete path.
Example: Manufacturing Stages
A component passes an initial inspection with probability
\[ 0.92. \]Given that it passes the initial inspection, it passes a stress test with probability
\[ 0.95. \]Given that it passes both earlier stages, it passes final calibration with probability
\[ 0.98. \]The probability of passing all three stages is
\[ 0.92(0.95)(0.98) = \boxed{0.85652}. \]Failure at a Later Stage
Using the same manufacturing example, suppose we want the probability that a component passes the initial inspection and stress test but fails final calibration.
The final failure probability is
\[ 1-0.98=0.02. \]Therefore,
\[ P(\text{pass, pass, fail}) = 0.92(0.95)(0.02) = \boxed{0.01748}. \]Multiplication and Complements
Complements frequently appear inside multiplication paths.
If
\[ P(B\mid A)=0.70, \]then
\[ P(B^c\mid A)=1-0.70=0.30. \]Therefore,
\[ P(A\cap B^c) = P(A)P(B^c\mid A). \]At Least One Versus All
Multiplication rules are especially direct when a problem asks for all stages to satisfy a condition.
By contrast, probabilities such as at least one often become easier through complements.
For example, if three independent trials each succeed with probability \(0.80\), then the probability of at least one failure is
\[ 1-P(\text{all three succeed}). \]Thus,
\[ 1-0.80^3 = \boxed{0.488}. \]Example: No Failures in Independent Trials
Suppose a device has probability
\[ 0.02 \]of failing during a test, independently from one device to another.
The probability that one device does not fail is
\[ 0.98. \]For five independent devices,
\[ P(\text{no failures}) = 0.98^5 \approx \boxed{0.9039}. \]Example: At Least One Failure
Continuing the previous example,
\[ P(\text{at least one failure}) = 1-P(\text{no failures}). \]Therefore,
\[ P(\text{at least one failure}) = 1-0.98^5 \approx \boxed{0.0961}. \]Multiplication for Exactly One Specified Path
Suppose three independent trials each have success probability
\[ p=0.70. \]The probability of the particular sequence
\[ SFS \]is
\[ 0.70(0.30)(0.70) = \boxed{0.147}. \]This is the probability of one specific ordered path.
A Specific Path Is Not the Same as “Exactly Two Successes”
In three trials, exactly two successes can occur through several paths:
\[ SSF,\qquad SFS,\qquad FSS. \]Multiplication gives the probability of each path.
Combining multiple paths requires addition after the path probabilities are calculated.
This lesson focuses on multiplication along paths; broader event aggregation will be developed further in later probability work.
Tree Principle: Multiply Along, Add Across
A useful organizing principle for probability trees is:
\[ \boxed{ \text{multiply along a path} }. \]If several mutually exclusive paths represent the same final event, their path probabilities can then be added:
\[ \boxed{ \text{add across alternative paths} }. \]The next lesson on total probability develops this aggregation principle in a systematic way.
Example: Employee Screening Path
Suppose \(70\%\) of applicants pass an initial screen:
\[ P(S)=0.70. \]Of those who pass, \(60\%\) pass an interview:
\[ P(I\mid S)=0.60. \]Of those who pass both, \(80\%\) accept an offer:
\[ P(A\mid S\cap I)=0.80. \]Then
\[ P(S\cap I\cap A) = 0.70(0.60)(0.80) = \boxed{0.336}. \]Thus \(33.6\%\) of applicants follow this complete path.
Reading Multiplication Problems from Language
Common phrases include:
- both \(A\) and \(B\);
- \(A\) followed by \(B\);
- first \(A\), then \(B\);
- all of the following occur;
- success on every trial;
- draw one type and then another;
- pass each stage;
- complete the entire sequence.
These phrases often indicate an intersection or path probability.
Write the Events Before Multiplying
Suppose a problem asks: “What is the probability that a customer renews and then upgrades?”
Let \(R\) mean renews and \(U\) mean upgrades.
The mathematical event is
\[ R\cap U. \]The general rule is
\[ P(R\cap U)=P(R)P(U\mid R). \]Writing the symbolic structure first helps prevent incorrect assumptions of independence.
Do Not Assume Independence Unless It Is Justified
The formula
\[ P(A\cap B)=P(A)P(B) \]is not the general multiplication rule.
It is the independent-events version.
The always-relevant structural formula is
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]Example: Why Independence Matters
A bag contains \(6\) green tokens and \(4\) yellow tokens. Two tokens are drawn without replacement.
A student incorrectly writes
\[ \frac{6}{10}\cdot\frac{6}{10}. \]But after one green token is removed, only \(5\) green tokens remain among \(9\) total tokens.
The correct probability is
\[ \frac{6}{10}\cdot\frac{5}{9} = \boxed{\frac{1}{3}}. \]Example: Conditional Information May Simplify a Problem
Suppose
\[ P(A)=0.50 \]and
\[ P(B\mid A)=0.30. \]No separate value of \(P(B)\) is needed to find the intersection:
\[ P(A\cap B) = 0.50(0.30) = \boxed{0.15}. \]The multiplication rule uses exactly the probabilities along the chosen path.
Recovering a Conditional Probability
If
\[ P(A\cap B) \]and
\[ P(A) \]are known, then the multiplication rule can be solved for \(P(B\mid A)\):
\[ P(B\mid A) = \frac{P(A\cap B)}{P(A)}. \]This returns the conditional probability formula.
Recovering a Missing Stage Probability
Suppose
\[ P(A)=0.80 \]and
\[ P(A\cap B)=0.52. \]Then
\[ 0.52=0.80P(B\mid A). \]Therefore,
\[ P(B\mid A) = \frac{0.52}{0.80} = \boxed{0.65}. \]Multiplication Rules and Percentages
Percentages may be multiplied after converting them to decimal or fractional form.
Suppose \(45\%\) of customers open a message, and \(20\%\) of those who open it click a link.
Then
\[ 0.45(0.20)=0.09. \]Therefore,
\[ \boxed{9\%} \]of all customers both open the message and click the link.
Conditional Percentages Are Relative to Their Parent Group
In the previous example, the \(20\%\) click rate is not \(20\%\) of all customers.
It is \(20\%\) of those who opened the message.
Multiplication converts that conditional rate into a joint rate relative to the full population:
\[ 45\%\times20\%=9\%. \]Multiplication Rules in Reliability
Suppose two components must both operate for a simple system to function.
If component \(A\) operates with probability
\[ 0.98 \]and component \(B\) operates independently with probability
\[ 0.97, \]then the probability both operate is
\[ 0.98(0.97)=\boxed{0.9506}. \]This calculation depends on the stated independence assumption.
Multiplication Rules in Finance
Suppose \(90\%\) of approved applications reach closing, and among those that close, \(96\%\) fund successfully.
Then the probability that an approved application both closes and funds is
\[ 0.90(0.96)=\boxed{0.864}. \]Thus \(86.4\%\) of approved applications follow the complete close-and-fund path under the stated model.
Multiplication Rules in Diagnostics
Diagnostic problems often involve quantities such as
\[ P(D) \]and
\[ P(+\mid D). \]Their product gives the joint probability
\[ P(D\cap +) = P(D)P(+\mid D). \]Later lessons will combine such joint probabilities using total probability and Bayes' theorem.
Multiplication Rules and Bayesian Reasoning
Bayes' theorem depends on connecting conditional probabilities to joint probabilities.
The multiplication identity
\[ P(A\cap B)=P(A)P(B\mid A) \]provides one direction.
Reversing the order gives
\[ P(A\cap B)=P(B)P(A\mid B). \]Equating these two representations is one of the key steps behind Bayes' theorem.
Common Error: Multiplying Marginal Probabilities for Dependent Events
A common mistake is to write
\[ P(A\cap B)=P(A)P(B) \]without checking independence.
The correct general form is
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]Common Error: Forgetting to Update After Sampling Without Replacement
Suppose a box contains \(4\) red and \(6\) blue objects.
If one red object is drawn and not replaced, the next draw occurs from
\[ 3\text{ red}+6\text{ blue}=9\text{ objects}. \]The second red probability is therefore
\[ \frac{3}{9}, \]not
\[ \frac{4}{10}. \]Common Error: Adding Along a Single Path
A complete sequential path represents an intersection.
Intersections along a path are calculated by multiplication, not addition.
If
\[ P(A)=0.60 \]and
\[ P(B\mid A)=0.50, \]then
\[ P(A\cap B)=0.60(0.50)=0.30, \]not
\[ 0.60+0.50. \]Common Error: Multiplying Alternative Paths Without Reason
Distinct complete paths in a tree usually represent mutually exclusive alternatives.
Multiplication is used within each path.
If several paths contribute to the same event, their completed path probabilities are generally combined by addition, not multiplication.
Common Error: Confusing \(P(A\cap B)\) with \(P(B\mid A)\)
The joint probability
\[ P(A\cap B) \]is measured relative to the entire sample space.
The conditional probability
\[ P(B\mid A) \]is measured within \(A\).
They are connected by multiplication:
\[ P(A\cap B)=P(A)P(B\mid A). \]Common Error: Ignoring Order in Conditional Factors
For a sequence \(A\) then \(B\), the natural factorization is
\[ P(A)P(B\mid A). \]Writing
\[ P(A)P(A\mid B) \]mixes incompatible pieces and does not generally produce the desired intersection.
Follow the actual path of information.
Common Error: Treating “At Least One” as a Single Multiplication Path
“At least one” usually describes several possible paths.
For repeated independent trials, the complement is often simpler:
\[ P(\text{at least one success}) = 1-P(\text{no successes}). \]Multiplication is then used to calculate the probability of the single complement path when appropriate.
A Reliable Procedure for Two Events
- Identify the intersection or sequence being requested.
- Choose a starting event.
- Write its probability.
- Write the probability of the next event conditioned on what has already occurred.
- Multiply the probabilities along the path.
- Use unconditional factors only when independence justifies them.
- Check that the result lies between \(0\) and \(1\).
A Reliable Procedure for Several Stages
- Write the events in their actual sequence.
- Start with the probability of the first event.
- For each later stage, condition on the path already completed.
- Update counts or probabilities when the process changes the state.
- Multiply all branch probabilities along the chosen path.
- If several paths are required, calculate each path separately before combining them.
- Interpret the final value relative to the original sample space.
Example: Full Independent Workflow
A server completes a task successfully with probability
\[ 0.99 \]on each of four independent jobs.
Find the probability that all four jobs succeed.
Independence allows the same unconditional probability at each stage:
\[ P(\text{all four}) = 0.99^4. \]Therefore,
\[ \boxed{ P(\text{all four})\approx0.9606 }. \]Example: Full Dependent Workflow
A container holds \(10\) components, \(7\) acceptable and \(3\) defective. Three components are selected without replacement.
Find the probability that all three selected components are acceptable.
First acceptable:
\[ \frac{7}{10}. \]Second acceptable given the first:
\[ \frac{6}{9}. \]Third acceptable given the first two:
\[ \frac{5}{8}. \]Therefore,
\[ P(\text{all acceptable}) = \frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}. \]Simplifying,
\[ \boxed{ P(\text{all acceptable})=\frac{7}{24} }. \]Practice 1: General Multiplication Rule
Suppose
\[ P(A)=0.60 \]and
\[ P(B\mid A)=0.40. \]Find \(P(A\cap B)\).
Practice 2: Independent Events
Suppose \(A\) and \(B\) are independent, with
\[ P(A)=0.70,\qquad P(B)=0.30. \]Find \(P(A\cap B)\).
Practice 3: Two Dice
Two fair dice are rolled independently. Find the probability that the first die shows \(6\) and the second die shows an even number.
Practice 4: Without Replacement
A bag contains \(8\) red tokens and \(2\) blue tokens. Two tokens are drawn without replacement. Find the probability that both are red.
Practice 5: With Replacement
Repeat Practice 4, but replace the first token before drawing the second. Find the probability that both are red.
Practice 6: Three-Stage Path
Suppose
\[ P(A)=0.80, \] \[ P(B\mid A)=0.75, \]and
\[ P(C\mid A\cap B)=0.90. \]Find \(P(A\cap B\cap C)\).
Practice 7: Complement Along a Path
Suppose
\[ P(A)=0.65 \]and
\[ P(B\mid A)=0.80. \]Find \(P(A\cap B^c)\).
Practice 8: Recover a Conditional Probability
Suppose
\[ P(A)=0.50 \]and
\[ P(A\cap B)=0.20. \]Find \(P(B\mid A)\).
Practice 9: Three Draws
A box contains \(4\) white balls and \(6\) black balls. Three balls are drawn without replacement. Find the probability of drawing white, then black, then white.
Practice 10: Identify the Error
A deck contains \(52\) cards. Two cards are drawn without replacement. A student writes
\[ P(\text{two kings}) = \frac{4}{52}\cdot\frac{4}{52}. \]Explain the error and give the correct multiplication expression.
Practice Answers
Practice 1
\[ P(A\cap B) = P(A)P(B\mid A) = 0.60(0.40) = \boxed{0.24}. \]Practice 2
Because the events are independent,
\[ P(A\cap B) = P(A)P(B) = 0.70(0.30) = \boxed{0.21}. \]Practice 3
The first die shows \(6\) with probability
\[ \frac{1}{6}, \]and the second die is even with probability
\[ \frac{3}{6}=\frac{1}{2}. \]Therefore,
\[ \boxed{ \frac{1}{6}\cdot\frac{1}{2} = \frac{1}{12} }. \]Practice 4
Without replacement,
\[ P(\text{both red}) = \frac{8}{10}\cdot\frac{7}{9}. \]Thus,
\[ \boxed{ P(\text{both red})=\frac{28}{45} }. \]Practice 5
With replacement, the draws are independent:
\[ P(\text{both red}) = \frac{8}{10}\cdot\frac{8}{10} = \boxed{\frac{16}{25}}. \]Practice 6
\[ P(A\cap B\cap C) = 0.80(0.75)(0.90) = \boxed{0.54}. \]Practice 7
First,
\[ P(B^c\mid A)=1-0.80=0.20. \]Therefore,
\[ P(A\cap B^c) = 0.65(0.20) = \boxed{0.13}. \]Practice 8
\[ P(B\mid A) = \frac{P(A\cap B)}{P(A)} = \frac{0.20}{0.50} = \boxed{0.40}. \]Practice 9
White first:
\[ \frac{4}{10}. \]Then black:
\[ \frac{6}{9}. \]Then white:
\[ \frac{3}{8}. \]Therefore,
\[ P(WBW) = \frac{4}{10}\cdot\frac{6}{9}\cdot\frac{3}{8} = \boxed{\frac{1}{10}}. \]Practice 10
The first king is not replaced, so after a king is drawn only \(3\) kings remain among \(51\) cards.
The correct expression is
\[ \boxed{ P(\text{two kings}) = \frac{4}{52}\cdot\frac{3}{51} }. \]What to Remember
The general multiplication rule is
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]Equivalently,
\[ \boxed{ P(A\cap B)=P(B)P(A\mid B) }. \]For independent events,
\[ \boxed{ P(A\cap B)=P(A)P(B) }. \]- Intersections and complete paths are calculated by multiplication.
- Dependent events require conditional probabilities.
- Independent events allow unconditional probabilities to be multiplied.
- Without-replacement sampling usually requires updating probabilities.
- For several stages, condition each new stage on the path already completed.
- In trees, multiply along a path and combine alternative paths only afterward.
The core habit is: follow the path and update each probability using the information already accumulated.
Where We Go Next
Multiplication rules calculate the probability of individual intersections and sequential paths. Many problems, however, reach the same event through several mutually exclusive paths.
The next lesson, Total Probability, develops a systematic method for partitioning the sample space, calculating the probability along each path, and combining those paths into an overall probability.