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

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**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.

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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:
\[
P(\text{Same Color}) = \frac{2}{5} + \frac{1}{15} = \frac{6}{15} + \frac{1}{15} = \frac{7}{15}
\]

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###

Dalam sebuah kotak terdapat 4 bola merah dan 2 bola putih, diambil 2 bola satu persatu tanpa dikembalikan. Tentukan peluang terambil: a. Berturut-turut bola merah dan putih. b. Kedua bola berwarna sama.

This problem involves calculating the probability of drawing two balls (one red and one white, or two of the same color) from a set of 4 red balls and 2 white balls, without replacement. The solution involves conditional probability and combinatorics, suitable for students in Grades 10-12.

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