If a is the optimistic time, b is the pessimistic time and m is most likely time of an activity, the expected time of the activity, is A. $$\frac{{{\text{a}} + {\text{m}} + {\text{b}}}}{6}$$ B. $$\frac{{{\text{a}} + 2{\text{m}} + {\text{b}}}}{6}$$ C. $$\frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6}$$ D. $$\frac{{{\text{a}} + 5{\text{m}} + {\text{b}}}}{6}$$

The correct answer is $\frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6}$.

The expected time of an activity is the weighted average of the optimistic time, the most likely time, and the pessimistic time. The weights are 1/6, 4/6, and 1/6, respectively. This is because the optimistic time is the least likely time to occur, the most likely time is the most likely time to occur, and the pessimistic time is the most likely time to occur.

The formula for the expected time of an activity is:

$E = \frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6}$

where:

• $a$ is the optimistic time
• $m$ is the most likely time
• $b$ is the pessimistic time

For example, if the optimistic time for an activity is 1 day, the most likely time is 3 days, and the pessimistic time is 5 days, then the expected time of the activity is:

$E = \frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6} = \frac{{1 + 4 \times 3 + 5}}{6} = 3$ days