Let's go through the problem step-by-step to find the probability of drawing a black marble after one black marble is drawn and not replaced:
1. Initial Counts of Marbles:
- Black marbles: 4
- Red marbles: 2
- White marbles: 6
2. Total Initial Marbles:
The total number of marbles in the bag initially is:
[tex]\[
4 \ (\text{black}) + 2 \ (\text{red}) + 6 \ (\text{white}) = 12 \ \text{marbles}
\][/tex]
3. One Black Marble is Drawn and Not Replaced:
After drawing one black marble, the number of black marbles decreases by 1:
- Black marbles left: \(4 - 1 = 3\)
Also, the total number of marbles decreases by 1:
- Total marbles left: \(12 - 1 = 11\)
4. Calculating the Probability:
The probability of then drawing another black marble is the number of black marbles left divided by the total number of marbles left:
[tex]\[
\text{Probability} = \frac{\text{Number of black marbles left}}{\text{Total number of marbles left}} = \frac{3}{11}
\][/tex]
Thus, the probability of drawing a black marble next, after one has already been drawn and not replaced, is:
[tex]\[
\boxed{\frac{3}{11}}
\][/tex]
So, the correct answer is (C) [tex]\(\frac{3}{11}\)[/tex].
