Conditional Probability Word Problem: Flight Luggage Problem

Conditional Probability 11th-12th Grade
PROBLEM
Leah is flying from Moncton to Vancouver with a connection in Montreal. The probability that her first flight leaves on time is 0.1. If the flight is on time, the probability that her luggage will make the connecting flight in Montreal is 0.8, but if the first flight is delayed, the probability that the luggage will make it is only 0.6. What is the probability that Leah's luggage arrives in Vancouver with her?

What You Will Learn

  • How to identify conditional probability problems where later events depend on earlier outcomes
  • When and how to apply the law of total probability to find overall probabilities
  • How to construct and interpret probability tree diagrams for multi-stage processes
  • The distinction between conditional probability P(B|A) and independent events
  • How to verify probability answers using the multiplication and addition rules

Picture This

Leah is flying from Moncton to Vancouver with a connection in Montreal. The probability that her first flight leaves...

This tree diagram shows all possible paths for Leah's luggage. The key insight is that there are two different routes to success, each with its own probability that must be calculated separately and then combined.

Solution: Method 1 — The Law of Total Probability

This is a classic conditional probability problem where we need to find the overall probability by considering all possible scenarios that lead to our desired outcome.

Step 1 — Identify the paths to success

Leah's luggage can arrive with her in exactly two ways:

  • Path 1: Flight is on time AND luggage makes the connection
  • Path 2: Flight is delayed AND luggage makes the connection

Step 2 — Calculate the probability of Path 1

For the on-time scenario:

P(luggage arrives via on-time flight) = P(on time) × P(luggage makes it | on time) = 0.1 × 0.8 = 0.08

Step 3 — Calculate the probability of Path 2

For the delayed scenario:

P(luggage arrives via delayed flight) = P(delayed) × P(luggage makes it | delayed) = 0.9 × 0.6 = 0.54

Note that P(delayed) = 1 - P(on time) = 1 - 0.1 = 0.9.

Step 4 — Apply the law of total probability

Since these paths are mutually exclusive (the flight is either on time OR delayed, not both), we add their probabilities:

P(luggage arrives) = P(Path 1) + P(Path 2) = 0.08 + 0.54 = 0.62

Solution: Method 2 — Organized Table Approach

Another way to see this problem is through a systematic breakdown of all possible outcomes and their probabilities.

Step 1 — Set up the probability table

Let's organize all four possible combinations of flight timing and luggage outcome:

Flight StatusP(Flight Status)Luggage OutcomeP(Luggage | Flight Status)Joint Probability
On Time0.1Makes It0.80.1 × 0.8 = 0.08
On Time0.1Misses0.20.1 × 0.2 = 0.02
Delayed0.9Makes It0.60.9 × 0.6 = 0.54
Delayed0.9Misses0.40.9 × 0.4 = 0.36

Step 2 — Identify successful outcomes

The highlighted rows represent scenarios where the luggage "Makes It" to Vancouver. We want the probability of either of these successful outcomes.

Step 3 — Sum the joint probabilities

P(luggage arrives) = 0.08 + 0.54 = 0.62

As a verification, notice that all four joint probabilities sum to 1.00, as they should: 0.08 + 0.02 + 0.54 + 0.36 = 1.00

The probability that Leah's luggage arrives in Vancouver with her is 0.62 or 62%.

Verification

Let's check our answer using the fundamental rules of probability:

Verify that all probabilities are valid

  • All conditional probabilities are between 0 and 1 ✓
  • P(on time) + P(delayed) = 0.1 + 0.9 = 1.0 ✓
  • For each flight status, the luggage probabilities sum to 1:
    • On time: 0.8 + 0.2 = 1.0 ✓
    • Delayed: 0.6 + 0.4 = 1.0 ✓

Verify our answer makes sense

Our answer of 0.62 should be between the two conditional probabilities (0.6 and 0.8), but closer to 0.6 since delays are much more likely than on-time flights (0.9 vs 0.1). Indeed:

0.6 < 0.62 < 0.8 ✓

The answer is much closer to 0.6, which makes sense given that 90% of the probability mass comes from the delayed-flight scenario.

Double-check by calculating P(luggage misses)

The probability the luggage misses should be 1 - 0.62 = 0.38. Let's verify:

P(luggage misses) = 0.02 + 0.36 = 0.38 ✓

Watch Out For These

✗ Mistake 1: Using only one scenario
Some students calculate just 0.1 × 0.8 = 0.08 and stop there. This only accounts for the on-time scenario and ignores the much more likely delayed-flight path to success.
✗ Mistake 2: Treating events as independent
Calculating P(flight) × P(luggage) = 0.1 × 0.8 = 0.08 ignores that luggage success depends heavily on flight timing. You can't use just one conditional probability when they're different.
✗ Mistake 3: Adding conditional probabilities directly
Computing 0.8 + 0.6 = 1.4 makes no sense – these are conditional probabilities that apply to different scenarios, not outcomes you can simply add together.
✗ Mistake 4: Forgetting that P(delayed) = 0.9
Since P(on time) = 0.1, then P(delayed) must be 0.9, not 0.1. This is a common reading error that leads to calculating 0.1 × 0.6 = 0.06 instead of 0.9 × 0.6 = 0.54.

The Pattern Behind This

This problem exemplifies the Law of Total Probability, one of the fundamental tools in probability theory. The general formula when an event A can occur through multiple mutually exclusive scenarios B₁, B₂, ..., Bₙ is:

P(A) = P(A|B₁) × P(B₁) + P(A|B₂) × P(B₂) + ... + P(A|Bₙ) × P(Bₙ)

In our specific case:

  • A = "luggage arrives"
  • B₁ = "flight on time", B₂ = "flight delayed"
  • P(A|B₁) = 0.8, P(A|B₂) = 0.6
  • P(B₁) = 0.1, P(B₂) = 0.9

So: P(A) = 0.8 × 0.1 + 0.6 × 0.9 = 0.08 + 0.54 = 0.62

Key insight: When outcomes depend on earlier events, you cannot use simple probability multiplication. You must consider all possible paths and weight them by how likely each path is to occur.

This pattern appears everywhere: medical diagnosis (symptoms depend on disease), quality control (defect rates depend on production conditions), weather forecasting (rain probability depends on cloud types), and many more real-world scenarios.

How to Spot This Problem Type

Look for these tell-tale signs that signal a conditional probability problem requiring the law of total probability:

  • "If... then..." language: "If the flight is on time, then..." or "Given that..., the probability is..."
  • Multiple stages: The problem describes a sequence of events where later outcomes depend on earlier results
  • Different probabilities for different scenarios: The same event has different probabilities depending on what happened earlier
  • Conditional probability notation: Expressions like P(A|B) appearing in the problem or solution
  • Two-way dependence: Event B affects the probability of event A, and you need the overall probability of A
Don't confuse with independent events! If the problem states that events are independent, or if later probabilities don't change based on earlier outcomes, then this method doesn't apply. Independent events use simple multiplication: P(A and B) = P(A) × P(B).

Common contexts include: medical testing, manufacturing quality, transportation systems, communication networks, and any multi-step process where earlier failures affect later success rates.

Real Applications

This type of conditional probability calculation is essential in many professional contexts:

  • Medical diagnosis: Disease probability depends on test results, but test accuracy varies with the disease being present or absent. Overall diagnosis probability requires considering both positive and negative test scenarios weighted by disease prevalence.
  • Network reliability: Data packet success rates depend on network congestion levels. System performance analysts calculate overall delivery rates by considering peak and off-peak scenarios with their respective success probabilities.
  • Financial risk assessment: Investment returns depend on market conditions. Portfolio managers calculate expected returns by weighting bull-market and bear-market scenarios by their likelihood and respective return rates.
  • Quality control: Product defect rates vary with production conditions (temperature, humidity, shift timing). Overall defect probability considers all conditions weighted by how often each occurs.

What If?

1Better On-Time Performance
What if Leah's airline improved and now has a 0.3 probability of on-time departure? The conditional probabilities remain the same: 0.8 if on time, 0.6 if delayed. What's the new probability her luggage arrives?
Step 1 — Identify the new scenario

P(on time) = 0.3, so P(delayed) = 1 - 0.3 = 0.7. Conditional probabilities unchanged.

Step 2 — Calculate on-time path

P(arrives via on-time) = 0.3 × 0.8 = 0.24

Step 3 — Calculate delayed path

P(arrives via delayed) = 0.7 × 0.6 = 0.42

Step 4 — Apply total probability

P(luggage arrives) = 0.24 + 0.42 = 0.66

Step 5 — Verify

Answer 0.66 is between 0.6 and 0.8, and closer to 0.6 since delays (70%) still outweigh on-time flights (30%). Answer: 0.66 or 66%

2Three-City Journey
Leah now flies Moncton→Montreal→Toronto→Vancouver. First flight on-time probability is 0.2. If on time, Toronto connection probability is 0.9. If delayed, Toronto connection probability is 0.7. What's the probability her luggage reaches Vancouver?
Step 1 — Set up the scenarios

P(first flight on time) = 0.2, P(first flight delayed) = 0.8

Step 2 — On-time path probability

P(arrives via on-time) = 0.2 × 0.9 = 0.18

Step 3 — Delayed path probability

P(arrives via delayed) = 0.8 × 0.7 = 0.56

Step 4 — Total probability

P(luggage arrives) = 0.18 + 0.56 = 0.74

Step 5 — Verify

0.74 lies between 0.7 and 0.9, weighted toward 0.7 since delays are more common. Answer: 0.74 or 74%

3Working Backwards
Airlines report that overall luggage arrival probability is 0.75. If P(on time) = 0.2 and P(luggage makes it | on time) = 0.9, what must P(luggage makes it | delayed) be?
Step 1 — Set up the equation

Using total probability: 0.75 = 0.2 × 0.9 + 0.8 × P(makes it | delayed)

Step 2 — Calculate known term

0.2 × 0.9 = 0.18

Step 3 — Solve for unknown

0.75 = 0.18 + 0.8 × P(makes it | delayed)
0.57 = 0.8 × P(makes it | delayed)

Step 4 — Find the answer

P(makes it | delayed) = 0.57 ÷ 0.8 = 0.7125

Step 5 — Verify

Check: 0.2 × 0.9 + 0.8 × 0.7125 = 0.18 + 0.57 = 0.75Answer: 0.7125 or 71.25%

4Reverse Conditional
Given that Leah's luggage DID arrive in Vancouver (using original problem data), what's the probability that her first flight was actually on time? This requires Bayes' theorem.
Step 1 — Identify what we want

We want P(on time | luggage arrived). This is a Bayes' theorem problem.

Step 2 — Apply Bayes' theorem

P(on time | arrived) = P(arrived | on time) × P(on time) ÷ P(arrived)

Step 3 — Substitute known values

From original problem: P(arrived | on time) = 0.8, P(on time) = 0.1, P(arrived) = 0.62

Step 4 — Calculate

P(on time | arrived) = (0.8 × 0.1) ÷ 0.62 = 0.08 ÷ 0.62 ≈ 0.129

Step 5 — Interpret

Even when luggage arrives successfully, there's only a 12.9% chance the flight was on time, since delays are so much more common. Answer: 0.129 or 12.9%

Frequently Asked Questions

How do you calculate probability when outcomes depend on previous events? +
Use conditional probability with the law of total probability. First, identify all possible paths to the desired outcome. Then multiply probabilities along each path and add them together. In this problem, luggage can arrive via an on-time flight (0.1 × 0.8 = 0.08) or a delayed flight (0.9 × 0.6 = 0.54), giving total probability 0.08 + 0.54 = 0.62.
What is the difference between conditional and independent probability? +
In conditional probability, P(B|A) depends on whether A occurred. In independent probability, P(B|A) = P(B) regardless of A. Here, luggage transfer depends heavily on flight timing - P(luggage makes it | on time) = 0.8 while P(luggage makes it | delayed) = 0.6. If they were independent, both would equal the same value.
When do you use a probability tree diagram? +
Use tree diagrams for sequential events where later outcomes depend on earlier ones. Each branch represents a possible outcome with its probability. Multiply along branches to get joint probabilities, then add branches leading to your target event. This problem has two stages (flight timing, then luggage transfer) making a tree diagram ideal for visualization.
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Dr. Neven Jurkovic

Mathematics educator with expertise in probability theory and statistical reasoning. Passionate about making conditional probability intuitive through real-world applications.

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Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-08-27