AND and OR

Let's begin with OR. Using the relative frequency mathematics already developed, what is the probability of the event "5"? There are only four outcomes, so the event "5" probability is 4/36 or 1/9.

To make it more interesting, let's say that event "5" is the desired result of one of two study teams observing the same project problem in "dice rolling." One team we call study team 5; the other we call study team 7. We need to complete the study in the time to complete 36 rolls. Either team finishing first is okay since both are working on the same problem. The probability that study team 5 will finish its observations on time is 4/36. If any result other than "5" is obtained, more time will be needed.

Now let's expand the event to a "5" or a "7". Event "7" is our second study team, with on-time results if it observes a "7". Recall that there are six event "7" opportunities. There are ten outcomes between the two study teams that fill the bill, giving either a "5" or a "7" as the sum of the two faces. Then, by definition, the probability of the event "5 OR 7" is 10/36. The probability that one or the other team will finish on time is 10/36, which is less risky than depending solely on one team to finish on time. Note that 10/36 is the sum of the probabilities of event "7" plus event "5": 6/36 + 4/36 = 10/36. ^[2]

It is axiomatic in the mathematics of probability that if two events are mutually exclusive, then when one occurs the other will not:

p(A OR B) = p(A) + p(B)

Now let's discuss AND. Consider the probability of the schedule event "rolling a 6 on one face AND rolling a 1 on the other face." The probability of rolling a "6" on one die is 1/6, as is the probability of rolling a "1" on the other. For the event to be true (that is, both a "6" and a "1" occur), both outcomes have to come up. Rolling the first die gives only a one-in-six chance of a "6"; there is another one-in-six chance that the second die will come up "1", altogether a one-in-six chance times a one-in-six chance, or 1/36. Of course 1/36 is very much less than 1/6. If the outcomes above were about our study teams finishing on time, one observing for a "6" on one die and the other observing for a "1" on the other die, then the probability of both finishing on time is the product of their individual probabilities. Requiring them both to finish on time increases the risk of a joint on-time finish event. We will study this "merge point" event is more detail in the chapter on schedules.

Another axiom we can write down is that if two events are independent, in no way dependent on each other, then:

p(A AND B) = p(A * B) = p(A) * p(B)