To solve this problem, we need to calculate the probability of drawing 2 sharpened pencils in succession without replacement from a total of 12 pencils (consisting of 4 sharpened and 8 unsharpened pencils).
Let's break down the problem into two steps - one for each draw.
### Step 1: Probability of the First Draw
- Total pencils: 12 (4 sharpened + 8 unsharpened).
- Sharpened pencils: 4
The probability of drawing a sharpened pencil on the first draw is the ratio of the number of sharpened pencils to the total number of pencils:
[tex]\[
P(\text{First sharpened}) = \frac{\text{Number of sharpened pencils}}{\text{Total number of pencils}} = \frac{4}{12} = \frac{1}{3}
\][/tex]
### Step 2: Probability of the Second Draw
After drawing one sharpened pencil, there are 11 pencils left in the box and 3 of them are sharpened.
- Total pencils left: 11
- Sharpened pencils left: 3
The probability of drawing another sharpened pencil on the second draw, given that the first pencil drawn was sharpened, is:
[tex]\[
P(\text{Second sharpened | First sharpened}) = \frac{\text{Number of remaining sharpened pencils}}{\text{Total remaining pencils}} = \frac{3}{11}
\][/tex]
### Total Probability
Since these are two independent events that must both happen, we multiply the probabilities of both events:
[tex]\[
P(\text{Two sharpened pencils}) = P(\text{First sharpened}) \times P(\text{Second sharpened | First sharpened}) = \frac{1}{3} \times \frac{3}{11}
\][/tex]
Simplifying this multiplication:
[tex]\[
P(\text{Two sharpened pencils}) = \frac{1 \times 3}{3 \times 11} = \frac{3}{33} = \frac{1}{11}
\][/tex]
So, the correct probability that both pencils drawn will be sharpened is:
[tex]\[
\boxed{\frac{1}{11}}
\][/tex]
