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Let's break down the problem.

We have 11 counters in total:
- 5 yellow counters
- 6 purple counters

Bronwyn takes a counter at random, replaces it, and then takes a second counter. We need to:

### Part a) Draw a tree diagram
For this problem, I will describe the tree diagram structure.

1. **First Draw (Two Outcomes):**
   - Draw the first branch splitting into two options: "Yellow" and "Purple".
     - Probability of drawing yellow: \( \frac{5}{11} \)
     - Probability of drawing purple: \( \frac{6}{11} \)

2. **Second Draw (Two Outcomes for Each First Outcome):**
   - If the first draw was yellow:
     - Probability of drawing yellow again: \( \frac{5}{11} \)
     - Probability of drawing purple: \( \frac{6}{11} \)
   - If the first draw was purple:
     - Probability of drawing yellow: \( \frac{5}{11} \)
     - Probability of drawing purple: \( \frac{6}{11} \)

In summary, the tree diagram has four possible outcomes:
1. Yellow → Yellow: \( \frac{5}{11} \times \frac{5}{11} \)
2. Yellow → Purple: \( \frac{5}{11} \times \frac{6}{11} \)
3. Purple → Yellow: \( \frac{6}{11} \times \frac{5}{11} \)
4. Purple → Purple: \( \frac{6}{11} \times \frac{6}{11} \)

### Part b) Probability that at least one of the counters is yellow
To find the probability of at least one yellow counter, we can use complementary probability:
- The complement of "at least one yellow" is "no yellow," which means both counters are purple.
- The probability of drawing purple twice (no yellow at all) is:
  \[
  P(\text{Purple, Purple}) = \frac{6}{11} \times \frac{6}{11} = \frac{36}{121}
  \]
Thus, the probability of at least one yellow is:
\[
P(\text{At least one yellow}) = 1 - P(\text{Purple, Purple}) = 1 - \frac{36}{121} = \frac{85}{121}
\]

### Final Answer:
The probability that at least one of the counters Bronwyn takes is yellow is \( \frac{85}{121} \).

Would you like a detailed diagram, further clarifications, or help with another part? Here are some related questions:

1. How would the probabilities change if Bronwyn didn't replace the counters?
2. Can we calculate the expected number of yellow counters in repeated trials?
3. What is the probability that both counters drawn are yellow?
4. How would the probability change if there were more counters of each color?
5. What if Bronwyn took three counters instead of two?

**Tip:** For probability problems, always look for ways to use complementary probabilities to simplify calculations.

There are 11 counters in a box, of which 5 are yellow and 6 are purple. Bronwyn takes a counter at random from the box, replaces it, and then takes a second counter. a) Draw a tree diagram to show all the possible outcomes. b) Work out the probability that at least one of the counters Bronwyn takes is yellow. Give your answer as a fraction in its simplest form.

Explore a probability problem involving 11 counters, including the use of a tree diagram to display possible outcomes and calculations of the probability that at least one of two draws is yellow. Perfect for Grades 7-10 students, this example also demonstrates complementary probability.

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