**11. Correct. The answer is true.** In the oil sample set, the engineer has six
samples which satisfy the condition: 31, 32, 33, 34, and 35. The sampling
process is without replacement, so the probability that the first element is
light oil is 5/15. The second element will be light oil with a probability of
4/14. Finally, the third element will be light oil with a probability of 3/13.
Accordingly, the probability that the three elements are light oil is the
product of three probabilities: (5/15) (4/14) (3/13) ≈ 0.0220 or 2.20%.
In set notation, we have that P_{1}(light oil)
∩ P_{2}(light oil) ∩ P_{3}(light oil) ≈
2.20%, where the subscript represents the number of the draw.