If an activity has its optimistic, most likely and pessimistic times as 2, 3 and 7 respectively, then its expected time and variance are respectively
A. 3.5 and $$\frac{5}{6}$$
B. 5 and $$\frac{{25}}{{36}}$$
C. 3.5 and $$\frac{{25}}{{36}}$$
D. 4 and $$\frac{5}{6}$$
Answer: Option C

But In CPM method is shown the longest duration that means answer should be Maximum time
expected time = [2 + 7 + 4(3) ]/6 = 3.5
Variance = [(tp - to)/6]^2
= [(7-2)/6]^2 = (5/6) ^ 2 = 25/36
so right option is C