A box contains two red balls and four blue balls. A ball is drawn at random from the box and then replaced…

1. Analyze the Question

We are asked to find the probability of drawing two red balls from a box containing a mix of red and blue balls. Crucially, the first ball is replaced before the second draw, which means the total number of balls remains the same for both draws.

2. Explain the Concept

This is a problem involving independent events. Because the first ball is replaced, the outcome of the first draw does not affect the outcome of the second draw.

The formula for the probability of two independent events ($A$ and $B$) occurring is:

3. Show the Solution

• Step 1: Determine the total number of balls. Total balls = 2 (red) + 4 (blue) = 6 balls.

• Step 2: Find the probability of drawing a red ball on the first draw ($P(R_1)$). $$P(R_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{2}{6} = \frac{1}{3}$$

• Step 3: Find the probability of drawing a red ball on the second draw ($P(R_2)$). Since the ball was replaced, the box still contains 2 red balls and 6 balls in total. $$P(R_2) = \frac{2}{6} = \frac{1}{3}$$

• Step 4: Calculate the combined probability. Since the events are independent: $$P(R_1 \text{ and } R_2) = P(R_1) \times P(R_2)$$ $$P = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$

4. Identify the Answer

Comparing our result to the options provided, the correct choice is D.

5. Exam Tip

Always look for the keyword "replaced" or "with replacement" in probability questions. If you see it, it means the denominator (total number of items) stays the same for every draw. If the question says "without replacement," the denominator will decrease by 1 for the second draw!