The image contains a probability question in Indonesian. Here’s a translation and breakdown of the problem:

Problem Statement:

1. In a box, there are 4 red balls and 2 white balls. Two balls are drawn one at a time without replacement. Determine the probability of the following events:

   a. The two balls drawn are one red and one white.

   b. The two balls drawn are both of the same color.

Let's solve each part.

Part a: Probability of One Red and One White Ball

To find the probability of drawing one red and one white ball in two draws without replacement, we have two cases:

1. First draw: Red, Second draw: White
2. First draw: White, Second draw: Red

Case 1: Red first, then White

• Probability of drawing a red ball first: $\frac{4}{6} = \frac{2}{3}$
• Probability of drawing a white ball second: $\frac{2}{5}$ (since one red ball is already drawn)

The probability for this case is: $P(\text{Red, then White}) = \frac{2}{3} \times \frac{2}{5} = \frac{4}{15}$

Case 2: White first, then Red

• Probability of drawing a white ball first: $\frac{2}{6} = \frac{1}{3}$
• Probability of drawing a red ball second: $\frac{4}{5}$ (since one white ball is already drawn)

The probability for this case is: $P(\text{White, then Red}) = \frac{1}{3} \times \frac{4}{5} = \frac{4}{15}$

Adding both cases, the total probability of drawing one red and one white is: $P(\text{One Red and One White}) = \frac{4}{15} + \frac{4}{15} = \frac{8}{15}$

Part b: Probability of Both Balls Being the Same Color

This scenario involves two cases as well:

1. Both balls are red.
2. Both balls are white.

Case 1: Both Balls Are Red

• Probability of drawing a red ball first: $\frac{4}{6} = \frac{2}{3}$
• Probability of drawing another red ball second: $\frac{3}{5}$ (since one red ball is already drawn)

The probability for this case is: $P(\text{Both Red}) = \frac{2}{3} \times \frac{3}{5} = \frac{6}{15} = \frac{2}{5}$

Case 2: Both Balls Are White

• Probability of drawing a white ball first: $\frac{2}{6} = \frac{1}{3}$
• Probability of drawing another white ball second: $\frac{1}{5}$ (since one white ball is already drawn)

The probability for this case is: $P(\text{Both White}) = \frac{1}{3} \times \frac{1}{5} = \frac{1}{15}$

Adding both cases, the total probability of drawing two balls of the same color is: $P(\text{Same Color}) = \frac{2}{5} + \frac{1}{15} = \frac{6}{15} + \frac{1}{15} = \frac{7}{15}$