ProLog

In our ongoing series on polling theory and statistical analysis, we pick up where we left off from our discussion of probability, and introduce 2 new concepts: conditional probability and independence.

Conditional Probability

We’ve already seen that probability is defined as a value between 0 and 1 that represents the relatively likelihood that a particular event will occur.  If we define a random variable X to denote the roll of a single fair die, and the event A to denote the outcome of an even number, then:

formulas courtesy of mathURL.com

since there are 6 possible outcomes, 3 of which yield an even number.

Conditional probabilities allow us to express the probabilities of one event, assuming another event has occurred.  For example, if we write Pr(B|A), we are describing the probability that event B occurred, given event A has occurred.  Using our original event A above, if we define event B as rolling a 2, then:

since there are 3 possible outcomes (given the roll was an even number), 1 of which is a 2.

Finally, we can calculate the probability both A and B will occur.  This is the intersection of the events A and B, and its probability is calculated as:

So the probability of A and B is the probability of A, times the probability of B given A.  In our single die example, we can calculate the probability that the die is 2 and that the die is even as:

This should make sense since there’s only 1 roll of the 6 possibilities that make both events true.

Dependence

Sometimes, the probably of an event given another event is the same as the probability of that event alone.  When:

we say that M and N are independent.  Returning to the probability of 2 intersecting events:

if and only if A and B are independent

This doesn’t hold for our original events A and B (Pr(B|A) = 1/3, Pr(B) = 1/6), but let’s define a second random variable Y which corresponds to the roll of a second fair die, with event M denoting that the second die is even, and event N denoting that the second die rolls a 2.  So the probability of rolling two 2’s is:

Here we used the reciprocity of the “if and only if” clause; in other words, the rule can be reversed if necessary.  Common sense tells us that the roll of the second die is in no way influenced by the roll of the first and therefor they are independent of each other.  As a result, we can simply multiply the individual probabilities.

Event Unions

Since we’re talking about the intersection of 2 events, we should probably cover the union of them as well.  Where the intersection denotes events A and B occurring, union denotes A or B occurring.  To find the probability of a union of 2 events, the equation is:

To illustrate this relationship, consider a Venn diagram of events A and B:

If we simply added Pr(A) and Pr(B), we’ll include the area Pr(AB) twice, so we we need to subtract one out.  If we define event C as the value of the first die greater than or equal to 3, we can calculate:

Of the 6 possible outcomes, only rolling a 1 fails to satisfy the event A+B.  Draw a Venn diagram to convince yourself.

Finally, if Pr(AB) = 0, then Pr(A+B) = Pr(A) + Pr(B).  In this case, we call events A and B mutually exclusive.  Consider Pr(B+C).

Class dismissed.