Sure! Let's solve the problem step by step.
Bonnie has a total of 12 pencils in her pencil case: 4 sharpened and 8 unsharpened pencils. She randomly selects 2 pencils without replacement. We need to find the probability that both pencils will be sharpened.
### Step-by-Step Solution:
1. Total number of pencils:
[tex]\[
\text{Total pencils} = 4 \text{ (sharpened)} + 8 \text{ (unsharpened)} = 12
\][/tex]
2. First selection:
The probability that the first pencil Bonnie selects is sharpened:
[tex]\[
P(\text{first sharpened}) = \frac{4}{12}
\][/tex]
3. Second selection:
If the first pencil Bonnie selects is sharpened, then there are 3 sharpened pencils left out of a total of 11 remaining pencils. So, the probability that the second pencil is sharpened:
[tex]\[
P(\text{second sharpened | first sharpened}) = \frac{3}{11}
\][/tex]
4. Combined Probability:
To find the probability that both events occur, multiply the probabilities of each event:
[tex]\[
P(\text{both sharpened}) = P(\text{first sharpened}) \times P(\text{second sharpened | first sharpened})
\][/tex]
Plugging in the values:
[tex]\[
P(\text{both sharpened}) = \left( \frac{4}{12} \right) \times \left( \frac{3}{11} \right) = \frac{4 \times 3}{12 \times 11} = \frac{12}{132} = \frac{1}{11}
\][/tex]
### Conclusion:
The probability that both pencils Bonnie selects will be sharpened is:
[tex]\[
\boxed{\frac{1}{11}}
\][/tex]
So, the correct answer is C. [tex]$\frac{1}{11}$[/tex]
