Official Solution:
A = {7, 9, 15, 20}
B = {4, 5, 6, 11}
One number is randomly selected from each of the lists above to form list C. What is the probability that list C contains one number less than 10 and the other greater than 10?
A. \(\frac{3}{16}\)
B. \(\frac{3}{8}\)
C. \(\frac{7}{16}\)
D. \(\frac{1}{2}\)
E. \(\frac{5}{8}\)
The probability that the number selected from List A is less than 10 and the number from List B is greater than 10 is \(\frac{1}{2} * \frac{1}{4} = \frac{1}{8}\).
The probability that the number selected from List A is greater than 10 and the number from List B is less than 10 is \(\frac{1}{2} * \frac{3}{4} = \frac{3}{8}\).
Thus, the overall probability is \(\frac{1}{8} + \frac{3}{8} = \frac{1}{2}\).
Answer: D