From Branches to Conditional Probabilities
In Why \(P(A\mid B)\) Is Not \(P(B\mid A)\), the event after the bar determines the reference group. A tree diagram gives us another way to see that group: it lays out the possible paths of a multi-stage chance process. We can use the tree to find the probability of each complete path and then compare the paths that satisfy a condition.
As in Tree Diagrams for Multi-Stage Experiments, each branch represents a possible result at one stage. A branch after an earlier result is labeled with a probability that applies given that earlier result. To find the probability of a complete path, multiply the probabilities along that path. If an event can happen along more than one path, add the probabilities of those paths.
This path-based approach matches the conditional probability formula from The Conditional Probability Formula. The numerator is the probability of both events; the denominator is the probability of the condition. A tree helps organize those probabilities when the chance process has stages.
A condition may describe one branch or several complete paths. If the condition is one branch at the first stage, the reference group is everyone who follows that branch. If the condition is an outcome at the final stage, it may include paths that began with different first-stage results. In that case, include all of those paths in the denominator.
At each point in the tree, the probabilities of the branches leaving that point should add to 1. The probability of a complete path is not generally the probability written on just its last branch: that branch probability applies only among outcomes that reached that point. Multiplying along the path accounts for the earlier stages too.
Worked Example: Which Warehouse Sent a Delayed Package?
Worked Example: Which Warehouse Sent a Delayed Package?
A fictional delivery company ships 70% of its packages from its North warehouse and 30% from its South warehouse. Among packages from the North warehouse, 4% are delayed. Among packages from the South warehouse, 10% are delayed. Find the probability that a randomly selected package came from the South warehouse, given that it was delayed.
State: Let \(S\) mean that a package came from the South warehouse, and let \(L\) mean that it was delayed. We want \(P(S\mid L)\), the probability of a South-warehouse origin among delayed packages.
Plan: The first-stage branches are the two warehouses. The second-stage branch probabilities are conditional on the warehouse. Multiply along each path to find its probability. Then use all delayed-package paths as the reference group: a delayed package could have come from either warehouse.
Do: From the North warehouse, the delayed probability is 0.04 and the not-delayed probability is \(1-0.04=0.96\). From the South warehouse, the corresponding probabilities are 0.10 and \(1-0.10=0.90\). The tree and path calculations are:
The delayed-package paths have total probability \(0.028+0.030=0.058\). Only the South-and-delayed path belongs in the numerator for \(P(S\mid L)\), and its probability is 0.030.
As a check, first find the overall probability of a delay from the two delayed paths: \(P(L)=0.70(0.04)+0.30(0.10)=0.058\). The probability of both a South origin and a delay is \(P(S\cap L)=0.30(0.10)=0.030\). The conditional probability formula gives \(0.030/0.058\approx0.5172\), the same result.
Conclude: In this fictional delivery process, about 51.7% of delayed packages came from the South warehouse. The answer is not the South warehouse’s 10% delay rate: that rate is the probability of a delay given a South origin, while the question asks for the probability of a South origin given a delay.
When a Condition Includes Several Paths
A useful tree skill is recognizing that a condition can combine paths. Suppose the condition is “at least one green ball was drawn.” That condition includes paths with a green ball on the first draw, a green ball on the second draw, or green balls on both draws. To calculate a conditional probability, include every path meeting the condition in the denominator, but count each complete path only once.
The next example also shows why branch probabilities can change from one stage to the next. When drawing without replacement, the contents of the container change after the first draw. Thus, the second-draw probabilities depend on the first result.
Worked Example: First Draw Is Green Given at Least One Green
Worked Example: First Draw Is Green Given at Least One Green
A bag contains 5 green balls and 3 gold balls. Two balls are drawn in order without replacement. Find the probability that the first ball is green, given that at least one of the two balls is green.
State: Let \(F\) mean that the first ball is green, and let \(G\) mean that at least one of the two balls is green. The target is \(P(F\mid G)\).
Plan: Build a two-stage tree. At the first draw, the probabilities are \(5/8\) for green and \(3/8\) for gold. For the second draw, use the number of each color remaining after the first result. Multiply along each path. The condition \(G\) includes every path except gold followed by gold.
Do: If the first ball is green, 4 green and 3 gold balls remain out of 7. If the first ball is gold, 5 green and 2 gold balls remain. The complete path probabilities are:
The condition “at least one green” includes the first three paths. Their probabilities sum to \(20/56+15/56+15/56=50/56\). The event that the first ball is green includes the first two paths, whose probabilities sum to \(20/56+15/56=35/56\). Therefore:
Check the denominator another way. The only path with no green ball is gold followed by gold, with probability \((3/8)(2/7)=6/56\). Thus \(P(G)=1-6/56=50/56\), matching the sum of the three paths that meet the condition. The numerator is \(P(F\cap G)=P(F)=5/8=35/56\), because a green first draw automatically guarantees at least one green ball. The ratio is again \((5/8)/(50/56)=0.70\).
Conclude: Given that at least one of the two drawn balls is green, the probability that the first ball was green is 0.70. The denominator includes three paths, not just the path \(GG\), because the condition is met whenever either draw is green.
Worked Example: Rain and Low Attendance
Worked Example: Rain and Low Attendance
A fictional outdoor community event has a 40% chance of rain. If it rains, the probability of low attendance is 70%. If it does not rain, the probability of low attendance is 20%. Find the probability of rain, given that attendance is low.
State: Let \(R\) mean that it rains and \(L\) mean that attendance is low. We want \(P(R\mid L)\), the probability of rain among events with low attendance.
Plan: Use rain and no rain as the first-stage branches, then use attendance as the second stage. Find both paths that end in low attendance. Their combined probability is the denominator; the rain-and-low path is the numerator.
Do: The chance of no rain is \(1-0.40=0.60\). If it rains, the probability of attendance that is not low is \(1-0.70=0.30\). If it does not rain, that probability is \(1-0.20=0.80\). Multiplying along the four paths gives:
The two paths with low attendance have total probability \(0.28+0.12=0.40\). Only the rain-and-low path is in both \(R\) and \(L\). Therefore:
As a check, all four path probabilities sum to \(0.28+0.12+0.12+0.48=1.00\). Also, \(P(R\cap L)=0.28\) and \(P(L)=0.40\), so the conditional probability formula gives \(0.28/0.40=0.70\).
Conclude: In this fictional model, the probability of rain among events with low attendance is 0.70. This does not mean that 70% of rainy events have low attendance; that probability was given as \(P(L\mid R)=0.70\). Here, we reversed the conditional and used both possible causes of low attendance in the denominator.
A Reliable Tree-Diagram Routine
State what each first-stage and later-stage result means. Read branch probabilities after the first stage as conditional probabilities given the earlier result.
At each point, include all possible next results. Their probabilities should add to 1. If a complement is needed, subtract the stated probability from 1.
Each complete path describes results at every stage. Multiply its branch probabilities to obtain the probability of that path.
The condition determines the denominator paths. The numerator includes only paths that satisfy both the target event and the condition.
Divide the numerator total by the condition total, provided that the condition has positive probability. Explain the result in context, naming the group represented by the condition.
Common Mistakes and AP Exam Tips
- Using only the last branch probability. A second-stage branch such as “0.10 delayed” applies only among packages that reached that warehouse branch. Multiply it by the first-stage probability to get the complete-path probability.
- Adding when paths are sequential. Multiply probabilities along one path. Add probabilities for different, non-overlapping paths that satisfy the event. In the urn example, “at least one green” requires adding three complete-path probabilities.
- Leaving a path out of the condition total. For \(P(A\mid B)\), the denominator includes every path where \(B\) occurs, including paths where \(A\) does not occur. The condition defines the reference group.
- Reversing the conditional. \(P(R\mid L)\) is not generally the same as \(P(L\mid R)\). State the reference group in words before calculating.
- Using an unchanged branch probability when the process changes. In sampling without replacement, the second-draw probabilities depend on the first draw. Label each later branch using the probabilities that apply at that point.
- Giving only a decimal. A full-credit interpretation names the target group and stays in context: “Among events with low attendance, the probability of rain is 0.70.”
A final check can catch many errors: branch probabilities from a point should total 1, and the probabilities of all complete paths should also total 1. For a conditional probability, the numerator cannot exceed the denominator, and the denominator must be positive.
Check Your Understanding
For each question, show which paths belong in the numerator and denominator before calculating.
- A fictional cafe gets 60% of its coffee beans from Supplier A and 40% from Supplier B. A package is marked late with probability 0.05 for A and 0.10 for B. Find the probability a late package came from B.
- A box contains 4 blue and 2 red tokens. Two are drawn without replacement. List the probability of each complete path.
- Using the box in Question 2, find the probability the first token is red, given that at least one token is red.
- A park has a 30% chance of a windy day. On windy days, a kite demonstration is canceled with probability 0.40; on nonwindy days, it is canceled with probability 0.10. Find the probability of wind given cancellation.
- In your own words, explain why a conditional probability’s denominator may require adding more than one path probability.