### Part (a): Completing the Tree Diagram
To complete the tree diagram, we need to denote all possible outcomes of the two dice rolls and their respective probabilities. Let’s start by noting the given probabilities:
- The probability of rolling a 6 ([tex]$P(\text{6})$[/tex]) is [tex]\(\frac{1}{5}\)[/tex].
- The probability of not rolling a 6 ([tex]$P(\text{not 6})$[/tex]) is [tex]\(1 - P(\text{6}) = 1 - \frac{1}{5} = \frac{4}{5}\)[/tex].
To visualize this in a tree diagram for two rolls, we consider the outcomes of each roll:
1. First Roll
- Probability of rolling a 6: [tex]\( \frac{1}{5} \)[/tex]
- Probability of not rolling a 6: [tex]\( \frac{4}{5} \)[/tex]
2. Second Roll (for each outcome from the first roll)
- If the first roll was a 6:
- Probability of rolling another 6: [tex]\( \frac{1}{5} \)[/tex]
- Probability of not rolling a 6: [tex]\( \frac{4}{5} \)[/tex]
- If the first roll was not a 6:
- Probability of rolling a 6: [tex]\( \frac{1}{5} \)[/tex]
- Probability of not rolling a 6: [tex]\( \frac{4}{5} \)[/tex]
Thus, the tree diagram will look like:
```
(start)
/ \
6 not 6
(1/5) (4/5)
/ \ / \
6 not 6 6 not 6
(1/5) (4/5) (1/5) (4/5)
```
Where each branch represents a possible outcome of rolling the dice and is labeled with its corresponding probability.
### Part (b): Probability of Rolling Exactly One Six
To find the probability of rolling exactly one six in two rolls, we consider the following scenarios:
1. Rolling a 6 on the first roll and not a 6 on the second roll.
2. Rolling not a 6 on the first roll and a 6 on the second roll.
We calculate the probabilities for each scenario:
1. First Scenario (6 on the first roll, not 6 on the second roll):
[tex]\[
P(\text{6, not 6}) = P(\text{6}) \times P(\text{not 6}) = \frac{1}{5} \times \frac{4}{5} = \frac{4}{25}
\][/tex]
2. Second Scenario (not 6 on the first roll, 6 on the second roll):
[tex]\[
P(\text{not 6, 6}) = P(\text{not 6}) \times P(\text{6}) = \frac{4}{5} \times \frac{1}{5} = \frac{4}{25}
\][/tex]
To find the total probability of rolling exactly one six, we sum the probabilities of the two scenarios:
[tex]\[
P(\text{exactly one 6}) = P(\text{6, not 6}) + P(\text{not 6, 6}) = \frac{4}{25} + \frac{4}{25} = \frac{8}{25}
\][/tex]
Therefore, the probability of rolling exactly one six in two rolls of the biased dice is [tex]\(\frac{8}{25}\)[/tex], which is approximately 0.32 or 32%.
