There are two passengers remaining on a city bus that has 3 stops left on its route. What is the probability that one of them exits at the last stop, given that they will exit at different stops?

To solve this problem, we need to find the probability that one of the two remaining passengers exits at the last stop, given that they will exit at different stops. Let's approach it step by step.

### Step 1: Understand the Problem
- There are 2 passengers and 3 stops left.
- The passengers exit at different stops, meaning each passenger will exit at one of the 3 stops, and no two passengers will exit at the same stop.
- We want to find the probability that one of these passengers exits at the last stop (stop 3).

### Step 2: Calculate the Total Number of Favorable Outcomes
Since the two passengers must exit at different stops, we can consider the different scenarios:

- Passenger 1 exits at stop 1, Passenger 2 exits at stop 2.
- Passenger 1 exits at stop 1, Passenger 2 exits at stop 3.
- Passenger 1 exits at stop 2, Passenger 2 exits at stop 1.
- Passenger 1 exits at stop 2, Passenger 2 exits at stop 3.
- Passenger 1 exits at stop 3, Passenger 2 exits at stop 1.
- Passenger 1 exits at stop 3, Passenger 2 exits at stop 2.

There are \(3 \times 2 = 6\) possible combinations where the two passengers exit at different stops.

### Step 3: Count the Favorable Outcomes
To satisfy the condition that one of the passengers exits at the last stop (stop 3), we look at the combinations where stop 3 is one of the stops used:

1. Passenger 1 exits at stop 3, Passenger 2 exits at stop 1.
2. Passenger 1 exits at stop 3, Passenger 2 exits at stop 2.
3. Passenger 1 exits at stop 1, Passenger 2 exits at stop 3.
4. Passenger 1 exits at stop 2, Passenger 2 exits at stop 3.

There are 4 favorable outcomes.

### Step 4: Calculate the Probability
The probability that one of the passengers exits at the last stop, given that they will exit at different stops, is the number of favorable outcomes divided by the total number of possible outcomes:

\[
P(\text{one exits at last stop}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{4}{6} = \frac{2}{3}
\]

### Conclusion
The probability that one of the two passengers exits at the last stop, given that they will exit at different stops, is **\(\frac{2}{3}\)** or approximately **66.7%**.

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Do you want details or have any questions?

Here are 5 related questions you might find interesting:
1. What is the probability that both passengers exit at consecutive stops?
2. How would the probability change if there were 4 stops left instead of 3?
3. What is the probability that the first passenger exits at the first stop and the second exits at the last stop?
4. How does this probability change if there are 3 passengers instead of 2?
5. What if the order in which the passengers exit is not considered—how does that affect the probability?

**Tip:** When dealing with conditional probabilities, always carefully count the favorable outcomes based on the given condition.

Explore the probability that one of two passengers exits at the last stop on a city bus route, given they exit at different stops. Includes step-by-step calculation and explanation. Suitable for Grades 9-12.

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