The Multiplication Rule for Two Events

Why the Multiplication Rule Matters

Many probability questions ask whether both of two events occur. For example, a record may belong to a particular group and also meet a specified condition, or a case may pass one stage and then pass another. The multiplication rule connects the probability of the first event with the probability of the second event given that the first has occurred.

The essential distinction is between a conditional probability and an unconditional one. If $A$ and $B$ are events, $P(A\mid B)$ describes the probability of $A$ among outcomes where $B$ has occurred. It is not generally the same as $P(A)$. This tutorial uses that distinction to calculate $P(A\cap B)$, the probability that both $A$ and $B$ occur.

Definition: For events $A$ and $B$, the intersection $A\cap B$ is the event that both $A$ and $B$ occur. If $P(B)>0$, the multiplication rule is $$ P(A\cap B)=P(A\mid B)P(B). $$ Equivalently, if $P(A)>0$, then $P(A\cap B)=P(B\mid A)P(A)$.

The rule requires the conditional probability to be paired with the probability of the event after the vertical bar. In $P(A\mid B)P(B)$, the condition is $B$, so the multiplier is $P(B)$. The two factors do not have to describe independent events.

Why the Rule Holds

When $P(B)>0$, conditional probability is defined by restricting attention to outcomes in $B$:

P(A\mid B)=\frac{P(A\cap B)}{P(B)}.

The outcomes counted in $A\cap B$ are precisely the outcomes in $B$ that also belong to $A$. Multiplying both sides of the definition by $P(B)$ gives the multiplication rule:

P(A\mid B)P(B) =\frac{P(A\cap B)}{P(B)}P(B) =P(A\cap B).

This is a rearrangement of the conditional-probability definition, not an additional assumption about how the events are related. If $P(B)=0$, $P(A\mid B)$ is not defined by this formula, so the stated conditional-probability version of the rule cannot be used. In that case $A\cap B$ is contained in an event of probability zero and therefore has probability zero.

A Reliable Calculation Process

A word problem often gives one probability for an initial group or stage and another probability that is explicitly conditional on it. Translate the wording before multiplying: identify the event after “given,” then multiply its conditional probability by the probability of that conditioning event.

Name the events.
Write down what $A$ and $B$ mean in the context, then identify which event is conditioned on.

Match the two factors.
For $P(A\mid B)$, use $P(B)$ as the other factor. Check that both quantities refer to the same population or process.

Calculate and interpret.
Multiply to obtain $P(A\cap B)$, then state that it is the probability of both events occurring.

Worked Example: Two Stages in a Hypothetical Application Review

Suppose a hypothetical review process has probability $0.80$ of an application passing an initial completeness check. Among applications that pass that check, the probability of passing a technical review is $0.75$. Find the probability that an application passes both stages. Assume the conditional probability given is for the technical review among applications that passed the initial check.

Method. Let $C$ be the event that the application passes the completeness check, and let $T$ be the event that it passes the technical review. The requested event is $C\cap T$. The given conditional probability is $P(T\mid C)=0.75$, and $P(C)=0.80$.

Apply the multiplication rule with $C$ as the conditioning event:

P(C\cap T)=P(T\mid C)P(C) =0.75(0.80) =0.60.

Thus the probability of passing both stages is $0.60$, or $60\%$. As a check, the joint probability is no larger than either factor: $0.60\leq 0.75$ and $0.60\leq 0.80$. The common mistake is to multiply $P(C)$ by an unconditional probability of passing the technical review; the problem supplies a probability conditional on passing the first check, so that is the factor to use.

Using Counts to Find a Joint Probability

Sometimes a problem supplies counts rather than probabilities. Counts can be used to find the conditional probability first, after which the multiplication rule applies. Make sure the denominator of a conditional proportion is the number of outcomes in the conditioning group, not the total population.

Worked Example: A Hypothetical Claims File Review

A hypothetical review selects one file at random from a group of $300$ claims. Of these files, $120$ concern a specified service category. Among those $120$ files, $90$ include all required supporting documents. Find the probability that a randomly selected file concerns the specified category and includes all required documents.

Method. Let $S$ be the event that a file concerns the specified service category, and $D$ the event that it includes all required documents. From the counts,

P(S)=\frac{120}{300}=0.40.

The conditional probability of complete documentation among files in the service category uses $120$, not $300$, as its denominator:

P(D\mid S)=\frac{90}{120}=0.75.

Now use the multiplication rule:

P(S\cap D)=P(D\mid S)P(S) =0.75(0.40) =0.30.

The answer is $0.30$, or $30\%$. The count provides a direct check: $90$ of the $300$ files meet both conditions, and $90/300=0.30$. A common denominator error is to calculate $P(D\mid S)$ as $90/300$. That ratio is already the joint probability $P(S\cap D)$, not the conditional probability within $S$.

Grouped Events and the Direction of Conditioning

A “grouped” event can be an event defined by membership in a category, such as a specified region, product type, or review class. If a conditional rate is reported within that group, the multiplication rule gives the probability that a randomly selected outcome both belongs to the group and has the specified feature. The group’s probability supplies the base rate; the within-group rate supplies the conditional factor.

Worked Example: A Hypothetical Regional Service Group

In a hypothetical portfolio, $35\%$ of service requests come from a particular region. Of the requests from that region, $80\%$ are resolved within two business days. Find the probability that a randomly selected request is from that region and is resolved within two business days.

Method. Let $R$ denote the event that a request is from the region, and $Q$ the event that it is resolved within two business days. The data state $P(R)=0.35$ and $P(Q\mid R)=0.80$. Therefore,

P(R\cap Q)=P(Q\mid R)P(R) =0.80(0.35) =0.28.

The requested probability is $0.28$, or $28\%$. The result is plausible because the joint event is contained in the regional event, so its probability cannot exceed $0.35$. It also cannot exceed the within-region rate $0.80$.

The conditioning direction matters. The information is “resolved within two days among requests from the region,” which is $P(Q\mid R)$, not $P(R\mid Q)$. Reversing those events changes the question and cannot be done by simply swapping the terms in the given conditional probability.

Common Mistake: Multiplying Unconditional Probabilities

A tempting but unjustified calculation is $P(A\cap B)=P(A)P(B)$. The multiplication rule does not state this. Its correct factorization uses a conditional probability: $P(A\cap B)=P(A\mid B)P(B)$, or the reversed conditional form when appropriate. Replacing $P(A\mid B)$ by $P(A)$ is valid only when the probabilities happen to be equal, as under independence; independence must be established or given, not assumed just because two events are described in sequence.

For example, if $P(B)=0.40$ and $P(A\mid B)=0.75$, then $P(A\cap B)=0.30$. If someone instead uses $P(A)P(B)$, a value for $P(A)$ is needed, and it may not equal $0.75$. The conditional probability describes the rate inside $B$, while $P(A)$ describes the rate across the full sample space. Confusing these denominators is the source of many incorrect answers.

Key takeaway: For “both $A$ and $B$,” multiply the probability of the conditioning event by the probability of the other event given that condition. Preserve the direction of the conditional probability, and do not assume the events are independent.

Check Your Understanding

For each problem, identify the conditioning event before calculating.

In a hypothetical inspection program, $68\%$ of items pass a visual screen. Among those that pass, $75\%$ pass a second measurement check. What is the probability an item passes both? (Answer: $0.5100$)
A randomly selected hypothetical account has probability $0.36$ of being in a specified account group. Within that group, the probability of receiving a notice is $0.55$. Find the probability that an account is in the group and receives a notice. (Answer: $0.1980$)
A hypothetical file set contains $360$ invoices, of which $90$ were paid late. Among the late invoices, $2/3$ include a follow-up notice. If one invoice is selected uniformly at random, find the probability that it was paid late and includes a follow-up notice. (Answer: $0.1667$)
For events $A$ and $B$, suppose $P(B)=0.30$ and $P(A\cap B)=0.18$. Find $P(A\mid B)$. (Answer: $0.6000$)
In a hypothetical service program, $40\%$ of cases are in group $G$. Of cases in $G$, $70\%$ meet a stated completion condition. Of the cases meeting that condition within $G$, $25\%$ also receive a follow-up. Find the probability a randomly selected case is in $G$, meets the condition, and receives a follow-up. (Answer: $0.0700$)

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