Operations Management and Strategic Management · Project Management, Monitoring and Control

PERT Three Time Estimates and Probability of Completion

Updated 10 October 2026 · Fact-checked

PERT handles uncertain activity times using three estimates: optimistic (a), most likely (m) and pessimistic (b). Expected time is (a + 4m + b) ÷ 6 and variance is ((b − a) ÷ 6)². Add expected times and variances along the critical path, then use Z = (Due date − Project expected time) ÷ project standard deviation to find the probability.

Understand PERT and Probability of Completion

PERT (Programme Evaluation and Review Technique) is a network method for projects where activity times are uncertain, such as research or new product launches. Instead of one fixed time, you take three estimates for each activity.

The optimistic time (a) is the shortest time if everything goes well. The most likely time (m) is the time you expect most often. The pessimistic time (b) is the longest time if things go wrong, excluding disasters. PERT assumes the times follow a beta distribution, which gives the weighted average with weight 4 on m.

From the three estimates you get an expected time (te) and a variance for every activity. Variance shows how uncertain the activity is. A wide gap between b and a means high uncertainty. You then find the critical path using the expected times, exactly as in CPM.

The project's expected duration is the sum of expected times on the critical path. The project variance is the sum of the variances of the critical-path activities only. This relies on the assumption that activity times are independent. Taking the square root of the project variance gives the project standard deviation.

Finally, you assume the project duration is approximately normally distributed. You convert the target date into a Z value and read the probability from the normal table. CPM differs: it uses one deterministic time per activity and focuses on time-cost trade-off, while PERT is probabilistic and focuses on time uncertainty.

Key rules to remember

Expected time: te = (a + 4m + b) ÷ 6; a = optimistic, m = most likely, b = pessimistic.
Activity variance: σ² = ((b − a) ÷ 6)²; Depends only on a and b, not on m.
Activity standard deviation: σ = (b − a) ÷ 6; Square root of the variance.
Project expected time: Te = Σ te of critical path activities; Critical path is the longest path using expected times.
Project variance: σp² = Σ σ² of critical path activities; Add variances, never standard deviations. Use critical path only.
Z value: Z = (Ts − Te) ÷ σp; Ts = scheduled or target time. Look up Z in the normal table for P(project finishes by Ts).
Reading the Z: Z = 0 gives probability 0.5; Ts above Te gives Z positive and probability above 50%; Ts below Te gives probability below 50%.

How to solve PERT and Probability of Completion questions

Use this order for any PERT probability question. Keep a small table of a, m, b, te and variance for every activity.

1Compute te = (a + 4m + b) ÷ 6 for every activity.
2Compute variance = ((b − a) ÷ 6)² for every activity.
3Draw the network and list all paths. Find the critical path using the te values (the longest path).
4Add te along the critical path to get the project expected time Te.
5Add the variances of critical-path activities only to get σp², then take the square root to get σp.
6Calculate Z = (Ts − Te) ÷ σp, where Ts is the target time.
7Read the probability from the normal table (given in the question or exam). If Z is negative, use 1 − P(|Z|).
8State the answer in words: the probability that the project is completed within Ts days is x%.

Quickest way: Table-first shortcut

When to use it: Use when the question gives a table of activities with predecessors and three estimates, and the marks are mostly for the final probability.

Write te in one column and variance in the next. Use (b − a)² ÷ 36 directly for variance.
Find the longest path by adding te along each path. Do not compute variances for non-critical paths.
Sum te and variance for the critical path only.
Compute σp and Z in one line, then use the table value.
If the critical path is close to another path in length, say so in one line, since that path may also affect completion.

Common mistakes in PERT and Probability of Completion

Adding variances of all activities in the network.
Students forget that only the critical path decides project duration.
Fix: Mark the critical path first and sum variances for those activities alone.
Adding standard deviations instead of variances.
Standard deviation is shown in the table and looks easy to add.
Fix: Add variances, then take the square root of the total.
Using (a + m + b) ÷ 3 or dividing by 4 in te.
Confusing the PERT weighting with a simple average.
Fix: Remember the weights 1, 4, 1 and divide by 6.
Using the most likely time m to find the critical path.
Students skip computing te.
Fix: Always find the critical path using expected times.
Getting the sign of Z wrong or reading the table incorrectly.
Mixing up Ts and Te in the numerator.
Fix: Z = target − expected. If target is below expected, Z is negative and probability is below 0.5.
Squaring (b − a) but forgetting to divide by 36, or dividing by 6 without squaring.
Mixing up the standard deviation and variance formulas.
Fix: Variance = (b − a)² ÷ 36. Standard deviation = (b − a) ÷ 6.

Worked examples

Example 1

A project has a single critical path of three activities. Activity A: a = 2, m = 4, b = 6 days. Activity B: a = 3, m = 6, b = 9 days. Activity C: a = 4, m = 7, b = 16 days. Find the expected project time, project standard deviation and Z for completion in 20 days.

Show the solution

A: te = (2 + 16 + 6) ÷ 6 = 24 ÷ 6 = 4 days. Variance = (4 ÷ 6)² = 16 ÷ 36 = 0.444.
B: te = (3 + 24 + 9) ÷ 6 = 36 ÷ 6 = 6 days. Variance = (6 ÷ 6)² = 1.
C: te = (4 + 28 + 16) ÷ 6 = 48 ÷ 6 = 8 days. Variance = (12 ÷ 6)² = 4.
Project expected time Te = 4 + 6 + 8 = 18 days.
Project variance = 0.444 + 1 + 4 = 5.444. Standard deviation = √5.444 = 2.333 days.
Z = (20 − 18) ÷ 2.333 = 0.857.

Answer: Expected time is 18 days, standard deviation is about 2.33 days and Z is about 0.86. From the normal table, the probability of finishing in 20 days is about 80%.

Example 2

Critical path activities of a project have expected times of 10, 12 and 8 days, with variances 1, 4 and 4. Find the probability that the project is completed in 27 days or less. Also state the probability of completion in 30 days (use Z = 1 → 0.8413 and Z = 0.67 → 0.7486 approx).

Show the solution

Project expected time Te = 10 + 12 + 8 = 30 days.
Project variance = 1 + 4 + 4 = 9. Standard deviation = 3 days.
For 27 days: Z = (27 − 30) ÷ 3 = −1.
P(Z ≤ −1) = 1 − P(Z ≤ 1) = 1 − 0.8413 = 0.1587.
For 30 days: Z = (30 − 30) ÷ 3 = 0, so probability = 0.5.

Answer: The probability of completing in 27 days or less is about 15.87%. The probability of completing in 30 days is 50%.

Exam tips

Show a neat table of a, m, b, te and variance. Step marks are given for each column.
Write the critical path clearly and state that only its variances are summed.
If the normal table is not given, state the Z value and the probability you use, and write the final answer in a sentence.
In theory questions on PERT versus CPM, write that PERT uses three time estimates and probability, while CPM uses a single time estimate and links time with cost.
Check Z sign before looking up the table. A target below expected time must give probability below 50%.

Practice questions from Project Management, Monitoring and Control

PERT and Probability of Completion in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CMA FinalPERT: Three Time Estimates and Probability

PERT and Probability of Completion: frequently asked questions

What is the difference between PERT and CPM?

PERT uses three time estimates and treats activity times as uncertain, so it gives probabilities of meeting deadlines. CPM uses one known time per activity and studies the trade-off between time and cost. PERT suits new, uncertain projects while CPM suits routine, well-understood ones.

How do I calculate expected time and variance in PERT?

Expected time is (a + 4m + b) ÷ 6. Variance is ((b − a) ÷ 6)². For the project, add the expected times and variances of the critical path activities.

Why do we use only the critical path for probability?

The critical path decides the project duration. The other paths have slack, so their delays do not affect completion unless they exceed that slack. The standard exam method therefore uses the critical path alone.

What if Z is negative?

A negative Z means the target date is earlier than the expected time. The probability is then less than 50%. Find the table value for the positive Z and subtract it from 1.