The probability that Ryan’s bus has to wait at the traffic lights in the morning on the way to school is 2/5.

Draw a probability tree to show the possibilities that the bus has to wait or can drive through the traffic lights on the two consecutive mornings, the bus

i. has to wait the lights on both occasions___________ ii. does not have to wait on either morning___________

iii has to wait on just one morning. ___________

To solve this problem, we need to draw a probability tree diagram that represents the possible outcomes for the bus having to wait or not wait at the traffic lights over two consecutive mornings. 

### Step-by-Step Explanation

1. **Define Probabilities:**
   - The probability that the bus has to wait at the traffic lights on any given morning is \( P(\text{Wait}) = \frac{2}{5} \).
   - The probability that the bus does not have to wait is \( P(\text{No Wait}) = 1 - P(\text{Wait}) = 1 - \frac{2}{5} = \frac{3}{5} \).

2. **Draw the Probability Tree:**
   - Start with a root node for the first morning. From this node, branch out to two possibilities: "Wait" with a probability of \( \frac{2}{5} \) and "No Wait" with a probability of \( \frac{3}{5} \).
   - From each of these branches, draw two more branches for the second morning, again with probabilities \( \frac{2}{5} \) for "Wait" and \( \frac{3}{5} \) for "No Wait".

3. **Calculate Combined Probabilities:**
   - To find the probability for each outcome over two mornings, multiply the probabilities along the branches.

Here is a description of the probability tree diagram:

#### Probability Tree Diagram

- **First Morning:**
  - **Wait** (Probability: \( \frac{2}{5} \))
    - **Second Morning:**
      - **Wait** (Probability: \( \frac{2}{5} \)) → Combined Probability: \( \frac{2}{5} \times \frac{2}{5} = \frac{4}{25} \)
      - **No Wait** (Probability: \( \frac{3}{5} \)) → Combined Probability: \( \frac{2}{5} \times \frac{3}{5} = \frac{6}{25} \)
  - **No Wait** (Probability: \( \frac{3}{5} \))
    - **Second Morning:**
      - **Wait** (Probability: \( \frac{2}{5} \)) → Combined Probability: \( \frac{3}{5} \times \frac{2}{5} = \frac{6}{25} \)
      - **No Wait** (Probability: \( \frac{3}{5} \)) → Combined Probability: \( \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \)

#### Answers to Questions

i.

Explore a probability tree diagram illustrating the chances that Ryan’s bus has to wait at traffic lights over two mornings. This problem involves calculating probabilities based on given conditions. Suitable for Grades 7-9 students learning probability theory.

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