When A and B are not independent, then one becomes a condition on the outcome of the other. For example, the question might be: What is the probability of A given the condition that B has occurred? [3]

Consider the situation where there are 12 marbles in a jar, 4 black and 8 white. Marbles are drawn from the jar one at a time, without replacement. The p(black marble on first draw) = 4/12. Then, a second draw is made. Conditions have changed because of the first draw. There are only 3 black marbles left and only 11 marbles in the jar. The p(black marble on the second draw given a black marble on the first draw) = 3/11, a slightly higher probability than 4/12 on the first draw. The probability of the second draw is conditioned on the results of the first draw.

The probability of a black marble on each of the first two draws is the AND of the probabilities of the first and second draw. We write the equation:

where B = event "black on the first draw," A = event "black on second draw, given black on first draw," and the notation "|" means "given." Filling in the numbers, we have:

Consider the project situation given in Figure 2-2. There we see two tasks, Task 1 and Task 2, with a finish-to-start precedence between them. [4] The project team estimates that the probability of Task 1 finishing at the latest finish of "on time + 10 days" is 0.45. Task 1 finishing is event A. The project team estimates that the probability of Task 2 finishing on time is 0.8 if Task 1 finishes on time, but diminishes to 0.4 if Task 1 finishes in "on time + 10 days." Task 2 finishing is event B. If event A is late, it overlaps the slack available to B. We calculate the probability of "A * B" as follows:

p(A10 * B0) = p(B0 | A10) * p(A10) = 0.4 * 0.45 = 0.18

"On Time" Latesl Finish, 10days

Event A = Task 1 Finish

Latesl Finish

Event B = Task 2 Finish

if A is (ale. it impacts the slack available lor B to se on tlnre (latest finish)

All probabilities are eslirnated oy (lie project team p(B ori lime, given A late as 10 > p(S not on time, given A late as late as 10 days) = 0 45 p{j4 not iaie as 10 days) = O.iiS

p(A lale as 10 days and B on time) = 0,4 * 0,45 = 0,18 Figure 2-2: Conditions in Task Schedules.

where A10 = event of Task 1 finishing "on time + 10 days" and B0 = event of Task 2 finishing on time if Task 1 finishes "on time + 10 days." Conclusion: there is less than one chance in five that both events A10 and B0 will happen. The project manager's risk management focus is on avoiding the outcome of B finishing late by managing Task 1 to less than "on time + 10 days."

There is a lot more to know about conditional probabilities because they are very useful to the project manager in decision making and planning. We will hold further discussion until Chapter 4, where conditional probabilities will be used in decision trees and tables.

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What you need to know aboutâ€¦ Project Management Made Easy! Project management consists of more than just a large building project and can encompass small projects as well. No matter what the size of your project, you need to have some sort of project management. How you manage your project has everything to do with its outcome.

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