Multiply When Independent Events Both Need to Occur
In The Definition \(P(A\mid B)=P(A)\), independence meant that learning one event occurred does not change the probability of the other. There is another useful way to express that same idea: when events \(A\) and \(B\) are independent, the probability that both occur is the product of their individual probabilities. This multiplication rule is useful for questions such as the chance that two traffic lights are both green.
The phrase “both occur” signals the intersection \(A\cap B\), also written \(A\text{ and }B\). Its probability is a joint probability, measured from the full chance process. The individual probabilities \(P(A)\) and \(P(B)\) describe each event separately; multiplying them gives the probability of the overlap only when the events are independent.
The rule works because independence says that knowing \(A\) occurred leaves the probability of \(B\) unchanged: \(P(B\mid A)=P(B)\), when \(P(A)>0\). The general multiplication rule from The General Multiplication Rule gives \(P(A\cap B)=P(A)P(B\mid A)\). Substituting \(P(B)\) for \(P(B\mid A)\) gives the independence multiplication rule.
This connection is important: multiplication is not automatically the right way to find the probability of “A and B.” For any events with \(P(A)>0\), the general rule uses the conditional probability \(P(B\mid A)\). The simpler product \(P(A)P(B)\) is appropriate when the events are independent. In this tutorial, independence is given or justified by the model; deciding whether events are independent from probability values is the subject of the next tutorial.
Using the Multiplication Rule
A careful solution starts by defining the events and checking what the question asks. If it asks for the chance that both events happen, identify the intersection. Then confirm that the events are independent, write the product, substitute the probabilities, and interpret the result in context.
State exactly what \(A\) and \(B\) mean in the chance process.
Use the problem’s stated assumption or a clear feature of the chance process. Do not assume independence just because two events are described separately.
For independent events, calculate \(P(A\cap B)=P(A)P(B)\).
Describe the chance that both named events occur, using the situation’s context.
A product between 0 and 1 is itself a probability. It cannot be greater than either of the two individual probabilities: the event that both occur is contained within each individual event. This provides a quick reasonableness check. If a calculated intersection probability exceeds \(P(A)\) or \(P(B)\), recheck the arithmetic or the setup.
Worked Example: Two Independent Traffic Lights
Worked Example: Two Independent Traffic Lights
Suppose a simplified traffic model treats the states of two lights on a route as independent. Let \(A\) be the event that the first light is green when a driver arrives, and \(B\) the event that the second light is green when the driver arrives. The model gives \(P(A)=0.60\) and \(P(B)=0.40\). Find the probability that both lights are green.
The question asks for both events, so the target is \(P(A\cap B)\). The model explicitly says the lights’ states are independent, so the multiplication rule applies.
As an arithmetic check, \(60\%\) of \(40\%\) is \(24\%\), which is \(0.24\). The result is also no greater than either individual probability: \(0.24\leq 0.40\leq 0.60\).
Conclusion: Under this independent-lights model, the probability that both lights are green when the driver arrives is \(0.24\), or 24%. This is a model-based probability, not a claim that traffic-light states in every real situation are independent.
Worked Example: Two Devices Pass Their Checks
Worked Example: Two Devices Pass Their Checks
A fictional equipment model treats the pass-or-fail results for two devices as independent. Let \(A\) mean device 1 passes its check, and let \(B\) mean device 2 passes its check. The model assigns probability \(0.96\) to device 1 passing and probability \(0.90\) to device 2 passing. What is the probability that both devices pass?
Because the model specifies that the two check results are independent, multiply the individual pass probabilities to find their intersection.
To check the multiplication, \(96\times90=8640\). There are four decimal places in the two factors combined, so \(0.8640=0.864\). The answer is below both \(0.96\) and \(0.90\), as a probability of both passing must be.
Conclusion: In this fictional independent-device model, the probability that both devices pass their checks is \(0.864\), or 86.4%. The result refers to both devices passing, not to at least one passing.
Worked Example: Both Shots Score
Worked Example: Both Shots Score
In a simplified sports model, a player takes two shots, and the outcomes are treated as independent. Let \(A\) mean the first shot scores and \(B\) mean the second shot scores. Suppose the probability of scoring on the first shot is \(0.70\), and the probability of scoring on the second is \(0.80\). Find the probability of scoring on both shots.
The events are defined for separate shots, and independence is an explicit assumption of this model. The desired outcome is \(A\cap B\), so use the product of the two scoring probabilities.
A second way to check is to note that \(7\times8=56\); the two factors have two decimal places altogether, giving \(0.56\). The answer is less than \(0.70\), the smaller individual probability, which is reasonable for an event requiring both shots to score.
Conclusion: Under the stated model, the probability that the player scores on both shots is \(0.56\), or 56%. The independence assumption matters: if scoring on the first shot changes the chance of scoring on the second, this product would not be justified using the individual probabilities alone.
What Independence Adds to “And”
The word “and” tells you that the question concerns an intersection, but it does not tell you which multiplication rule to use. If events are independent, multiply their unconditional probabilities. If they are not known to be independent, the general multiplication rule from The General Multiplication Rule uses a conditional probability instead:
That formula accounts for the chance of \(B\) after \(A\) is known to have occurred. Independence allows the conditional probability \(P(B\mid A)\) to be replaced by \(P(B)\). The definition in The Definition \(P(A\mid B)=P(A)\) expresses the same idea with the events in the other order: learning \(B\) does not change the probability of \(A\). Thus, the product rule is a consequence of independence, not a way to declare independence without support.
For example, if a problem gives \(P(A)=0.5\) and \(P(B)=0.4\), those two values alone do not determine \(P(A\cap B)\). The events might be independent, or the chance of \(B\) might differ among outcomes where \(A\) occurs. You need information about independence or about a conditional probability to calculate their intersection.
Also distinguish “both” from “either.” “Both \(A\) and \(B\)” means the intersection and, for independent events, calls for multiplication. “\(A\) or \(B\)” means the union and is handled using the addition rules from earlier tutorials. Do not use the multiplication rule to answer a union question.
Common Mistakes and AP Exam Tips
- Multiplying without checking independence. A statement that two events have been described separately does not establish independence. State the independence assumption or explain why the chance process supports it.
- Adding for “both.” The addition rules concern “or,” not “and.” For the intersection of independent events, multiply the two probabilities.
- Using the rule for “at least one.” The product \(P(A)P(B)\) gives the probability that both occur, not the probability that either or both occur. Read the requested event carefully.
- Confusing a conditional probability with an unconditional one. The general multiplication rule uses \(P(B\mid A)\). Replace it with \(P(B)\) only when independence is established.
- Reporting a number without identifying its meaning. A complete response says what the probability represents and names both events in context.
Key Takeaway
For two independent events, multiply their individual probabilities to find the probability that both occur. The condition of independence is essential: without it, use the general multiplication rule and the relevant conditional probability. In every solution, make clear which events are being multiplied and what their product means in context.
Check Your Understanding
For each situation, write the event or events clearly, show the calculation, and interpret the result in context.
- Two independent signals each have probability \(0.85\) of working. What is the probability that both signals work?
- A model says two independent traffic lights are green with probabilities \(0.30\) and \(0.50\). Find the probability that both are green.
- Two independent attempts have success probabilities \(0.60\) and \(0.75\). Find the probability that both attempts succeed. Check that your result is no greater than either individual probability.
- A problem gives \(P(A)=0.40\) and \(P(B)=0.70\), but does not state that the events are independent or provide a conditional probability. Can you determine \(P(A\cap B)\) from those values alone? Explain.
- For independent events \(C\) and \(D\), \(P(C)=0.25\) and \(P(D)=0.80\). A student adds the probabilities to find the chance both occur. Identify the error and calculate the correct probability.