Independence of Three or More Events

Extending the Product Rule to Several Events

In Unions with Dependent Events, you used conditional probability to find the overlap of two events when one could change the probability of the other. When events are independent, the product rule is simpler: multiply the probabilities of the events. This tutorial extends that rule to three or more events, as in a system that works only if every component passes its check.

For two independent events \(A\) and \(B\), the earlier tutorial The Multiplication Rule for Independent Events established that \(P(A\cap B)=P(A)P(B)\). For several events, the same idea applies when the events are mutually independent. Mutual independence is a stronger requirement than checking that each pair is independent.

Definition: Events \(A_1,A_2,\ldots,A_n\) are mutually independent if, for every selection of two or more of these events, the probability that all selected events occur equals the product of their individual probabilities. In particular, for three events \(A\), \(B\), and \(C\), mutual independence requires the product rule to hold for each pair and for all three together.

For three mutually independent events, the probability that all three occur is

P(A\cap B\cap C)=P(A)P(B)P(C).

For \(n\) mutually independent events, multiply the probability of each event to find the probability that they all occur:

P(A_1\cap A_2\cap\cdots\cap A_n) =P(A_1)P(A_2)\cdots P(A_n).

This rule is not a new shortcut that works for any collection of events. The events must be mutually independent. If knowing that one event occurred changes the probability of another, use the general multiplication rule from earlier in the course, with conditional probabilities, rather than automatically multiplying marginal probabilities.

Why Mutual Independence Matters

For three events, checking the three pairs alone does not establish mutual independence. Mutual independence requires the product rule to work for \(A\) and \(B\), for \(A\) and \(C\), for \(B\) and \(C\), and for the intersection of all three. For more events, the definition similarly requires the product rule for every group of two or more events.

In a probability model, independence should come from how the chance process works or be stated as an assumption. For example, separate component checks might reasonably be modeled as independent if one result does not affect the others and there is no shared factor influencing them. But if the checks share a power supply that can fail, their outcomes may be related. Multiplying probabilities is justified by the independence model, not just by the fact that several probabilities are provided.

Important distinction: Pairwise independence means every pair of events is independent. Mutual independence requires the product rule for every subcollection, including the collection of all the events. Pairwise independence alone is not enough to use the product rule for all events together.

Once mutual independence is established, the rule can also be applied to complements. For example, if a component passes with probability \(0.96\), its probability of failing is \(1-0.96=0.04\). If component outcomes are mutually independent, events describing failures or passes across different components can be combined with the product rule as appropriate.

A Plan for Several Independent Events

A good solution starts by defining the events precisely. “The system passes” may mean every component passes, while “at least one component fails” is a different event. Next, identify the independence assumption and match each probability to the event it describes. Then multiply for an intersection, or use a complement if that makes the requested event easier to calculate.

Define the events.
State what each event represents and describe the combined event being requested.

Check mutual independence.
Use the chance-process description or a stated model assumption. Do not assume pairwise independence is sufficient.

Choose the calculation.
For all events occurring, multiply their probabilities. For a complementary event, first find the probability of the simpler opposite outcome.

Interpret the result.
State what the probability means in the situation, and include the relevant components or checks.

A series system is a system that works only if every component works. So the probability that a series system works is the probability that all component-success events occur. If those events are mutually independent, multiply their success probabilities. The system fails if at least one component fails, which is the complement of all components working.

Worked Examples

Worked Example: Three Independent Component Checks

A device passes its final check only if three separate components each pass. Suppose a probability model assigns pass probabilities \(0.98\), \(0.95\), and \(0.90\) to components 1, 2, and 3, respectively, and assumes the three pass events are mutually independent. Find the probability that the device passes the final check.

State. Let \(A\), \(B\), and \(C\) represent the events that components 1, 2, and 3 pass. The final check passes when \(A\cap B\cap C\) occurs.

Plan. The model explicitly assumes mutual independence, so the product rule for three events applies. The device must pass all three checks; this is an intersection, not a union.

Do. Multiply the three component pass probabilities:

P(A\cap B\cap C) =P(A)P(B)P(C) =(0.98)(0.95)(0.90) =(0.931)(0.90) =0.8379.

The intermediate product is \(0.98(0.95)=0.931\), and \(0.931(0.90)=0.8379\). Thus, the probability is \(0.8379\), or \(83.79\%\).

Conclude. Under the stated mutual-independence model, the probability that all three components pass and the device passes its final check is \(0.8379\), or \(83.79\%\).

Worked Example: At Least One Failure in a Four-Component System

A monitoring unit works only if all four of its components pass. Suppose the pass probabilities for components 1 through 4 are \(0.90\), \(0.95\), \(0.98\), and \(0.99\), and the pass events are mutually independent. Find the probability that at least one component fails.

State. Let \(A_i\) be the event that component \(i\) passes, for \(i=1,2,3,4\). The target is the probability that at least one component fails.

Plan. The complement of “at least one fails” is “all four pass.” Since the pass events are mutually independent, multiply their probabilities to find the probability of all four passing, then subtract from 1. This complement method avoids listing the different ways one or more components could fail.

Do. First find the probability that every component passes:

P(A_1\cap A_2\cap A_3\cap A_4) =(0.90)(0.95)(0.98)(0.99) =(0.855)(0.98)(0.99) =(0.8379)(0.99) =0.829521.

Therefore, the probability of at least one failure is

1-P(A_1\cap A_2\cap A_3\cap A_4) =1-0.829521 =0.170479 \approx 0.1705.

The multiplication checks as \(0.855(0.98)=0.8379\), followed by \(0.8379(0.99)=0.829521\). The final probability is rounded to four decimal places.

Conclude. Under the stated independence model, the probability that at least one of the four components fails is approximately \(0.1705\), or \(17.05\%\).

Worked Example: Pairwise Independence Is Not Enough

Consider one outcome chosen at random from the four equally likely triples \(000\), \(011\), \(101\), and \(110\). Let \(A\) mean the first digit is 1, \(B\) mean the second digit is 1, and \(C\) mean the third digit is 1. Check whether the three events are mutually independent.

State. We need to determine whether the product rule holds for every pair and for all three events together.

Plan. There are four equally likely outcomes. Count how many satisfy each event and each relevant intersection. Pairwise independence requires each pair-intersection probability to equal the product of its marginal probabilities; mutual independence also requires the three-event intersection to equal the product of all three marginal probabilities.

Do. Each digit is 1 in two of the four outcomes, so

P(A)=P(B)=P(C)=\frac{2}{4}=\frac{1}{2}.

For each pair, exactly one outcome satisfies both events: \(A\cap B\) occurs in \(110\), \(A\cap C\) occurs in \(101\), and \(B\cap C\) occurs in \(011\). Thus, each pair-intersection probability is \(1/4\), which matches the product of its marginal probabilities:

P(A\cap B)=P(A)P(B)=\frac{1}{4}, \qquad P(A\cap C)=P(A)P(C)=\frac{1}{4}, \qquad P(B\cap C)=P(B)P(C)=\frac{1}{4}.

However, no listed outcome has all three digits equal to 1. Therefore, \(P(A\cap B\cap C)=0\). If the three events were mutually independent, their three-event intersection would have probability

P(A)P(B)P(C) =\left(\frac{1}{2}\right)^3 =\frac{1}{8}.

Since \(0\ne 1/8\), the product rule does not hold for all three events together.

Conclude. The events are pairwise independent, but they are not mutually independent. Pairwise checks alone do not justify multiplying all three probabilities.

Common Mistakes and AP Exam Tips

Multiplying without establishing independence. The product rule for several events requires mutual independence. Point to a stated assumption or explain why the chance process supports it.
Checking only the pairs. For three events, pairwise independence does not guarantee that all three are mutually independent. The all-three intersection must also satisfy the product rule.
Confusing “all” and “at least one.” “All components pass” is an intersection. “At least one fails” is its complement in a system where each component either passes or fails.
Using failure probabilities when the event requires success. Identify the event first. For a system that needs every component to pass, multiply the pass probabilities, not the failure probabilities.
Giving a probability without context. A complete conclusion names the event and the modeled situation, such as the chance that all four components pass.

AP Exam Tip: Define the events, state the mutual-independence assumption, show the product for the intersection, and interpret the result in context. If you use a complement, name the opposite event and show the subtraction from 1. Do not claim that pairwise independence alone makes the whole collection mutually independent.

Key Takeaway

For three or more mutually independent events, multiply the individual probabilities to find the probability that all of them occur. Mutual independence is stronger than pairwise independence, so check what the model or chance process actually supports. For “at least one” questions, consider whether the complement is easier to calculate.

Key takeaway: If \(A_1,\ldots,A_n\) are mutually independent, then \(P(A_1\cap\cdots\cap A_n)=P(A_1)\cdots P(A_n)\). Use this product only when mutual independence is justified.

Check Your Understanding

Show your reasoning and interpret probability answers in context.

Three independent alarms each have probability \(0.8\) of sounding during a test. What is the probability that all three sound?
Using the alarms in Question 1, find the probability that at least one does not sound. Show how the complement helps.
For three events, what intersection probabilities must satisfy the product rule for the events to be mutually independent?
Explain why finding that every pair in a group of three events is independent does not, by itself, prove mutual independence.
A series system works only if every component works. If its component pass events are mutually independent, which probability calculation gives the chance that the system works?