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To solve this problem, we need to calculate the probability that when a die with 2 blue sides and 4 red sides is thrown three times, it shows red twice and blue once.

### Step 1: Identify the probabilities of each outcome
- Probability of getting red on a single throw (\( P(R) \)) = \(\frac{4}{6} = \frac{2}{3}\)
- Probability of getting blue on a single throw (\( P(B) \)) = \(\frac{2}{6} = \frac{1}{3}\)

### Step 2: Calculate the probability of getting red twice and blue once
The desired outcome can happen in the following ways: RRB, RBR, BRR.

For each specific order (e.g., RRB):
\[ P(\text{RRB}) = P(R) \times P(R) \times P(B) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27} \]

Since there are 3 different arrangements (RRB, RBR, BRR):
\[ P(\text{2 Reds and 1 Blue}) = 3 \times \frac{4}{27} = \frac{12}{27} \]

### Step 3: Simplify the probability
\[ \frac{12}{27} = \frac{4}{9} \]

Thus, the correct answer is \(\frac{4}{9}\).

### Conclusion
The correct answer is the option labeled as "4/9".

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Would you like to go into more detail or have any questions?

Here are five follow-up questions that relate to this concept:
1. What is the probability of getting exactly one red and two blue outcomes?
2. How does the problem change if the die had 3 blue and 3 red sides?
3. What is the expected number of red sides in three throws?
4. How would the probability change if the die were thrown four times?
5. What if the die had different numbers of blue and red sides, such as 1 blue and 5 red?

**Tip:** When dealing with probability problems involving multiple outcomes, considering all possible arrangements can be crucial for finding the correct answer.

A die has 2 sides coloured blue and 4 sides coloured red. If the die is thrown three times, what is the probability that it will show red twice and blue once?

Learn how to calculate the probability of rolling a biased die (2 blue, 4 red) three times, showing red twice and blue once. This problem involves key probability concepts and the binomial probability theorem, suitable for students in Grades 9-11.

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