In a bag there are 6 balls of which 3 are white and 3 are black. They are drwn successively with replacement. What is the chance that the colours are alternate ?

Explanation

To solve the problem of finding the probability that the colors of the balls drawn from the bag are alternate, we can follow these steps:

### Step 1: Understand the Problem
We have a bag containing 6 balls: 3 white and 3 black. We will draw these balls successively with replacement. We need to find the probability that the colors of the balls drawn alternate.

### Step 2: Identify the Possible Patterns
The two possible alternating patterns for drawing 6 balls are:
1. White, Black, White, Black, White, Black (WBWBWB)
2. Black, White, Black, White, Black, White (BWBWBW)

### Step 3: Calculate the Probability for Each Pattern
Each time we draw a ball, the probability of drawing a white ball (W) is \( \frac{3}{6} = \frac{1}{2} \) and the probability of drawing a black ball (B) is also \( \frac{3}{6} = \frac{1}{2} \).

For the first pattern (WBWBWB):
- The probability of drawing in this order is:
\[ P(WBWBWB) = P(W) \times P(B) \times P(W) \times P(B) \times P(W) \times P(B) \]
\[ = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^6 = \frac{1}{64} \]

For the second pattern (BWBWBW):
- The probability of drawing in this order is:
\[ P(BWBWBW) = P(B) \times P(W) \times P(B) \times P(W) \times P(B) \times P(W) \]
\[ = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^6 = \frac{1}{64} \]

### Step 4: Combine the Probabilities
Now, we add the probabilities of both patterns:
\[ P(\text{Alternating colors}) = P(WBWBWB) + P(BWBWBW) = \frac{1}{64} + \frac{1}{64} = \frac{2}{64} = \frac{1}{32} \]

### Final Answer
The probability that the colors are alternate is \( \frac{1}{32} \).

---