If the optimistic time, most likely time and pessimistic time for activity A are 4, 6 and 8 respectively and for activity B are 5, 5.5 and 9 respectively, then A. expected time of activity A is greater than the expected time of activity B B. expected time of activity B is greater than the expected time of activity A C. expected time of both activities A and B are same D. none of the above

The correct answer is: B. expected time of activity B is greater than the expected time of activity A.

The expected time of an activity is calculated as follows:

$E(t) = \frac{(t_o + 4t_m + t_p)}{6}$

where $t_o$ is the optimistic time, $t_m$ is the most likely time, and $t_p$ is the pessimistic time.

For activity A, we have:

$E(t_A) = \frac{(4 + 4 \times 6 + 8)}{6} = 6$

For activity B, we have:

$E(t_B) = \frac{(5 + 4 \times 5.5 + 9)}{6} = 6.33$

Therefore, the expected time of activity B is greater than the expected time of activity A.

Here is a brief explanation of each option:

• Option A: expected time of activity A is greater than the expected time of activity B. This is incorrect, as shown above.
• Option B: expected time of activity B is greater than the expected time of activity A. This is correct, as shown above.
• Option C: expected time of both activities A and B are same. This is incorrect, as shown above.
• Option D: none of the above. This is incorrect, as option B is the correct answer.