Turn an Independence Check into a Context Sentence
In Independence and Real-World Assumptions, you considered whether a chance process gives a reason to treat events as independent. This tutorial focuses on explaining that decision with numbers. A good justification does more than state “independent” or “dependent”: it names the events, shows a relevant comparison, and explains what that comparison means in the situation.
Recall from The Definition \(P(A\mid B)=P(A)\) that, when \(P(B)>0\), events \(A\) and \(B\) are independent if learning that \(B\) occurred does not change the probability of \(A\). The comparison is between the conditional probability \(P(A\mid B)\) and the unconditional probability \(P(A)\). If they differ, knowing that \(B\) occurred changes the probability of \(A\), so the events are dependent.
There is another test from Testing Independence with the Product Rule. Compare the probability that both events occur with the product of their individual probabilities. This test is useful when the joint probability and the two marginal probabilities are supplied, or when a table gives the counts needed to calculate them.
The two comparisons express the same idea when \(P(B)>0\): by the conditional probability formula, \(P(A\mid B)=P(A\cap B)/P(B)\). Thus, the conditional comparison asks whether the probability changes after learning \(B\), while the product comparison asks whether the joint probability has the value independence would require.
A Reliable Pattern for Writing the Justification
Before calculating, define the events in words. Then identify which numerical comparison the information supports. If the conditional probability is supplied or can be calculated and the condition has positive probability, compare it with the matching unconditional probability. If the joint probability is available, compare it with the product of the two marginal probabilities.
Say what \(A\) and \(B\) mean, including the relevant people, setting, or time period.
Use \(P(A\mid B)\) versus \(P(A)\) only when \(P(B)>0\). Alternatively, compare \(P(A\cap B)\) with \(P(A)P(B)\).
Write the probabilities and calculations clearly, keeping the events in the same order in the comparison.
State whether the events are independent and explain what the numerical equality or difference says about the situation.
A useful dependent-events sentence is: “Because \(P(B\mid A)=\ldots\) differs from \(P(B)=\ldots\), knowing that [event \(A\)] occurred changes the probability of [event \(B\)]; therefore, the events are dependent in this model.” For an independent-events conclusion, say that the probabilities are equal and that knowing one event occurred does not change the probability of the other.
For the product-rule version, make the required value visible: “The probability of both events is \(\ldots\), while the product of their individual probabilities is \(\ldots\). Because these are [equal/not equal], the events are [independent/dependent] in this model.” A context sentence should make clear which situation the numbers describe, rather than leaving the comparison as an unexplained calculation.
Worked Examples
Worked Example: Bus Use and a Late Arrival
Suppose a hypothetical transportation model describes students arriving at a particular school on a particular morning. Let \(A\) be the event that a randomly selected student uses the bus, and let \(B\) be the event that the student arrives late. The model gives \(P(A)=0.40\), \(P(B)=0.20\), and \(P(B\mid A)=0.35\). Are \(A\) and \(B\) independent according to these probabilities?
State. We are checking whether bus use and arriving late are independent in this model. The supplied conditional probability is \(P(B\mid A)\), so we compare it with the unconditional probability \(P(B)\).
Plan. Since \(P(A)=0.40>0\), \(P(B\mid A)\) is defined. Independence would require \(P(B\mid A)=P(B)\). As a second check, we can find the joint probability using the general multiplication rule and compare it with the product of the marginal probabilities.
Do. The conditional probability is \(P(B\mid A)=0.35\), but the unconditional probability is \(P(B)=0.20\). These probabilities differ by \(0.35-0.20=0.15\), or 15 percentage points.
Using the general multiplication rule, the probability that a student both uses the bus and arrives late is
The multiplication checks as \(40\times35=1400\), with four decimal places in the factors, so the result is \(0.1400=0.14\). The value independence would require for the joint probability is
Here \(40\times20=800\), giving \(0.0800=0.08\). The actual joint probability in the model, \(0.14\), is not equal to the independence-rule value, \(0.08\).
Conclude. Bus use and arriving late are dependent in this model: among students who use the bus, the probability of arriving late is \(0.35\), compared with \(0.20\) for a randomly selected student overall. Knowing that a student uses the bus changes the modeled probability of arriving late. This conclusion applies to the specified model and morning; it does not establish why the probabilities differ.
Worked Example: A Coin and a Spinner
A classroom activity uses a fair coin and a spinner with eight equal sections, three of which are blue. Each is used once, and the result of one does not affect the other. Let \(A\) mean the coin lands heads, and let \(B\) mean the spinner lands on blue. Use the model’s probabilities to justify whether these events are independent.
State. The events are heads on the coin and blue on the spinner. We will compare the probability of both occurring with the product of the individual probabilities.
Plan. The fair coin gives \(P(A)=1/2\), and three of the spinner’s eight equal sections are blue, so \(P(B)=3/8\). The stated separate random mechanisms support the independence model. We will check numerically that its joint probability agrees with the product rule.
Do. There are two equally likely coin outcomes and eight equally likely spinner sections, giving \(2\times8=16\) equally likely outcome pairs. Heads and blue occur together in one coin outcome paired with each of the three blue sections, so there are three favorable pairs. Therefore,
The product of the marginal probabilities is
The two values are equal. A conditional check gives the same result: once the coin lands heads, the spinner still has three blue sections out of eight, so \(P(B\mid A)=3/8=P(B)\). The condition is valid because \(P(A)=1/2>0\).
Conclude. The events are independent in the stated model. The probability of heads and blue is \(3/16\), exactly the product of the probability of heads and the probability of blue. The separate random steps also give a reason that learning the coin result does not change the spinner’s chances.
Worked Example: Club Membership and Borrowing a Laptop
Consider a hypothetical roster of 200 students. A student is selected at random from the roster. Let \(A\) mean that the selected student belongs to a school club, and let \(B\) mean that the student borrowed a school laptop this week. Of the 200 students, 80 belong to a club, 90 borrowed a laptop, and 48 both belong to a club and borrowed a laptop. Are \(A\) and \(B\) independent for this random selection?
State. We are considering events for one randomly selected student from this roster. We will use the counts to compare the joint probability with the product of the marginal probabilities.
Plan. Each probability uses the full roster of 200 as its denominator. If the events are independent, \(P(A\cap B)\) must equal \(P(A)P(B)\). We can also compare the laptop-borrowing rate among club members with the overall laptop-borrowing rate.
Do. The individual and joint probabilities are
The product of the marginal probabilities is
The arithmetic is \(40\times45=1800\), giving \(0.1800=0.18\). Because \(0.24\ne0.18\), the joint probability is not the value required by independence.
For a conditional check, there are 48 laptop borrowers among the 80 club members, so
The comparison is \(P(B\mid A)=0.60\) versus \(P(B)=0.45\). The conditional rate is higher by \(0.60-0.45=0.15\), or 15 percentage points. Also, \(P(A)=0.40>0\), so this conditional probability is defined.
Conclude. Club membership and borrowing a laptop are dependent for a student selected at random from this roster: 60% of club members borrowed a laptop, compared with 45% of students overall. Knowing that the selected student belongs to a club changes the probability of having borrowed a laptop. This conclusion describes the roster and random selection, not necessarily students at other schools.
Worked Example: When One Event Has Probability Zero
In a hypothetical process, let \(A\) be the event that a randomly selected device is charged, with \(P(A)=0.70\). Let \(B\) be the event that the device has a model number from a category that does not exist in this process. Thus, \(P(B)=0\). Are \(A\) and \(B\) independent?
State. Event \(B\) is impossible in the specified process. We need to check independence without using a conditional probability whose condition has probability zero.
Plan. The conditional probability \(P(A\mid B)\) is not defined because \(P(B)=0\). Instead, use the product-rule criterion, which does not require a positive probability for \(B\), and compare \(P(A\cap B)\) with \(P(A)P(B)\).
Do. Since \(B\) cannot occur, \(A\cap B\) cannot occur either, so \(P(A\cap B)=0\). Also,
Thus, \(P(A\cap B)=P(A)P(B)\). The conditional comparison cannot be made: its formula would require division by \(P(B)=0\).
Conclude. The events are independent by the product-rule criterion, even though \(P(A\mid B)\) is undefined. This boundary case is why a positive-probability condition belongs with the conditional-probability test, not with the product-rule test.
Common Mistakes and What a Full-Credit Answer Says
- Giving only a label. “They are dependent” does not show how the probabilities support that conclusion. Include the comparison and connect it to what knowing one event does to the chance of the other.
- Comparing the wrong probabilities. If you use \(P(B\mid A)\), compare it with \(P(B)\), not with \(P(A)\). The event after the bar identifies the event whose probability is being assessed.
- Using a conditional probability when its condition has probability zero. If \(P(B)=0\), \(P(A\mid B)\) is undefined. Use \(P(A\cap B)=P(A)P(B)\) instead; do not divide by zero or call the conditional probability zero.
- Calling a small difference equality. Use unrounded values when possible. If displayed values are rounded, say that the comparison is approximate and use enough precision to support the conclusion.
- Mixing up percentage points and percent change. A change from \(0.45\) to \(0.60\) is a difference of 0.15, or 15 percentage points. Calling it “15 percent higher” would describe a different calculation.
- Overstating what a table proves. A conclusion based on a defined roster and random selection applies to that setting. Do not automatically extend it to a larger population that the information does not describe.
Key Takeaway
A clear independence justification is a short chain of reasoning: identify the events, compare probabilities using a valid criterion, and explain the comparison in context. The conditional test requires a positive probability for the condition; the product-rule test does not.
Check Your Understanding
For each question, show the relevant numerical comparison and write a conclusion in context.
- Suppose \(P(A)=0.30\), \(P(B)=0.50\), and \(P(B\mid A)=0.50\). What can you conclude about independence? What condition makes the conditional comparison valid?
- A model gives \(P(A)=0.40\), \(P(B)=0.25\), and \(P(A\cap B)=0.12\). Use the product rule to decide whether the events are independent.
- In a group of 100 randomly selected library users, 40 borrowed a tablet, 30 used a study room, and 18 did both. Define events for one randomly selected user and check independence using both the joint and conditional comparisons.
- Suppose \(P(B)=0\), \(P(A)=0.60\), and \(P(A\cap B)=0\). Is \(P(A\mid B)\) defined? What does the product-rule check show?
- Write a context sentence explaining why \(P(B\mid A)=0.42\) and \(P(B)=0.28\) indicate dependence, assuming \(P(A)>0\). Include the meaning of the comparison rather than only the label.