In an urn there are 4 white and 4 blank balls . What is the probalility of drawing the first ball white , the second black , the third white , and fourth black , and so on .

Answer

To solve the problem of finding the probability of drawing the first ball white, the second black, the third white, and the fourth black, and so on, we can follow these steps:

### Step 1: Understand the total number of balls
We have a total of 8 balls in the urn: 4 white balls and 4 black balls.

### Step 2: Calculate the probability of drawing the first ball (White)
The probability of drawing a white ball first (Event E1) is given by:
\[ P(E1) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{8} = \frac{1}{2} \]

### Step 3: Calculate the probability of drawing the second ball (Black)
After drawing one white ball, there are now 7 balls left (3 white and 4 black). The probability of drawing a black ball second (Event E2) is:
\[ P(E2) = \frac{\text{Number of black balls}}{\text{Total number of remaining balls}} = \frac{4}{7} \]

### Step 4: Calculate the probability of drawing the third ball (White)
Now, after drawing one white and one black ball, there are 6 balls left (3 white and 3 black). The probability of drawing a white ball third (Event E3) is:
\[ P(E3) = \frac{3}{6} = \frac{1}{2} \]

### Step 5: Calculate the probability of drawing the fourth ball (Black)
After drawing two balls (one white and one black), there are 5 balls left (3 white and 2 black). The probability of drawing a black ball fourth (Event E4) is:
\[ P(E4) = \frac{2}{5} \]

### Step 6: Continue this pattern
Following this pattern, we can calculate the probabilities for the subsequent draws:
- For the fifth ball (White):
\[ P(E5) = \frac{2}{4} = \frac{1}{2} \]
- For the sixth ball (Black):
\[ P(E6) = \frac{1}{3} \]
- For the seventh ball (White):
\[ P(E7) = \frac{1}{2} \]
- For the eighth ball (Black):
\[ P(E8) = \frac{0}{1} = 0 \quad (\text{This means we cannot draw a black ball anymore}) \]

### Step 7: Combine the probabilities
Now, we multiply all these probabilities together:
\[ P(E1) \times P(E2) \times P(E3) \times P(E4) \times P(E5) \times P(E6) \times P(E7) \times P(E8) = \left(\frac{4}{8}\right) \times \left(\frac{4}{7}\right) \times \left(\frac{3}{6}\right) \times \left(\frac{2}{5}\right) \times \left(\frac{2}{4}\right) \times \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right) \times 0 \]

Since one of the probabilities is zero, the overall probability is:
\[ P = 0 \]

### Final Answer
Thus, the probability of drawing the balls in the specified order (first white, second black, etc.) is **0**.

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