A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate colour.

Answer

To solve the problem of finding the probability that four balls drawn from a bag containing 4 red and 4 blue balls are in alternate colors, we can follow these steps:

### Step 1: Understand the Problem
We have a total of 8 balls: 4 red and 4 blue. We need to find the probability that when we draw 4 balls one by one, they alternate in color.

### Step 2: Identify the Two Possible Patterns
There are two patterns for drawing the balls in alternate colors:
1. Blue, Red, Blue, Red (BRBR)
2. Red, Blue, Red, Blue (RBRB)

### Step 3: Calculate the Probability for the First Pattern (BRBR)
1. **First Draw (Blue):** There are 4 blue balls out of 8 total balls.
\[ P(\text{Blue first}) = \frac{4}{8} = \frac{1}{2} \]

2. **Second Draw (Red):** After drawing a blue ball, there are 7 balls left (4 red, 3 blue).
\[ P(\text{Red second}) = \frac{4}{7} \]

3. **Third Draw (Blue):** After drawing one red ball, there are 6 balls left (3 red, 3 blue).
\[ P(\text{Blue third}) = \frac{3}{6} = \frac{1}{2} \]

4. **Fourth Draw (Red):** After drawing one blue and one red, there are 5 balls left (3 red, 2 blue).
\[ P(\text{Red fourth}) = \frac{3}{5} \]

5. **Total Probability for BRBR:**
\[ P(BRBR) = \frac{4}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{1}{2} \times \frac{4}{7} \times \frac{1}{2} \times \frac{3}{5} = \frac{6}{70} = \frac{3}{35} \]

### Step 4: Calculate the Probability for the Second Pattern (RBRB)
1. **First Draw (Red):** There are 4 red balls out of 8 total balls.
\[ P(\text{Red first}) = \frac{4}{8} = \frac{1}{2} \]

2. **Second Draw (Blue):** After drawing a red ball, there are 7 balls left (3 red, 4 blue).
\[ P(\text{Blue second}) = \frac{4}{7} \]

3. **Third Draw (Red):** After drawing one blue ball, there are 6 balls left (3 red, 3 blue).
\[ P(\text{Red third}) = \frac{3}{6} = \frac{1}{2} \]

4. **Fourth Draw (Blue):** After drawing one red and one blue, there are 5 balls left (2 red, 3 blue).
\[ P(\text{Blue fourth}) = \frac{3}{5} \]

5. **Total Probability for RBRB:**
\[ P(RBRB) = \frac{4}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{1}{2} \times \frac{4}{7} \times \frac{1}{2} \times \frac{3}{5} = \frac{6}{70} = \frac{3}{35} \]

### Step 5: Combine the Probabilities
Since both patterns are mutually exclusive, we can add the probabilities:
\[ P(\text{Alternate Colors}) = P(BRBR) + P(RBRB) = \frac{3}{35} + \frac{3}{35} = \frac{6}{35} \]

### Final Answer
The probability that the drawn balls are in alternate colors is:
\[ \frac{6}{35} \]

---