Basic Concepts of Probability
6.8 Conditional Probabilities
The probability of randomly selecting a student on campus who is both a psychology major and a biological female is 8%. Since there is a dependency between eventsAandB, we shouldnotuse Formula 6.5, and it would notbe true that the probability of randomly selecting a student on campus who is both a psychology major and a biological female is 6%.■
Formula 6.6 allows us to incorporate into the calculations the realization that most psychology majors are biological females. And so, of the popu- lation of students, the likelihood of event A and event B co-occurring in the same person is increased a bit by the fact that events A and Balready seem to co-occur a bit more than one might expect if the two events were randomly represented in the population (or said in other words, if the two events were independent of each other). Of course in other situations, events A and B may be dependent, but in such a way that the occurrence of A decreasesthe likelihood of eventBoccurring. The key term regarding whether events are independent of each other or not is whether the like- lihood of one event changes (increases or decreases) if the other event is known to have occurred.
and so on.) Determining the probability of rolling a die and getting a 4 changes depending on the number of sides a die has because the sample space in which a 4 can be found changes.
The formula for determining conditional probabilities is simply a rework- ing of the multiplication rule for two dependent events. The formula is as follows.
Conditional probability formula P A B =P A and B
P B (Formula 6.7)
By simply dividing both sides of the formula for the multiplication rule for two dependent events by P(B), we can transform Formula 6.6 into Formula 6.7.
Let us try our hand at determining conditionals by looking at a form of a problem that is often used when teaching probability–the ball-in-urn scenario.
Imagine an urn (it is a bit of a mystery why probability people like to call baskets
“urns,”but far be it for us to change the status quo!) filled with balls. These balls have different colors and also have either an“X”or a“Y”on them. Let us further stipulate that each ball is equally likely to be drawn. Now suppose the urn we are working with has the following contents:
20 Red–X
20 Red–Y
10 Green–X
50 Green–Y
Conveniently, the balls add up to 100. On the basis of the given information, we should be able to see the following:
P(Red) =.4 (Formula 6.1) P(Green) =.6 (Formula 6.1)
P(X) =.3 (Formula 6.1)
P(Y) =.7 (Formula 6.1)
P(Red or Green) =.6 + .4 = 1 (Formula 6.2) P(X or Y) =.3 + .7 = 1 (Formula 6.2) P(Red or X) =.4 + .3–.2 = .5 (Formula 6.4) P(Green or X) =.6 + .3–.1 = .8 (Formula 6.4) P(Red or Y) =.4 + .7–.2 = .9 (Formula 6.4) P(Green or Y) =.6 + .7–.5 = .8 (Formula 6.4)
Furthermore, we should be able to see that the following are true:
P(Red and X) =.2 (given) P(Red and Y) =.2 (given) P(Green and X) =.1 (given) P(Green and Y) =.5 (given)
Let us try to find some conditionals.
■ Question Given the probabilistic situation above, what are the following conditional probabilities? P(Red|X), P(Red|Y), P(Green|X), and P(Green|Y)?
Solution
ForP(Red|X), first assign“Red”and“X”to eventsAandB. Let us define“Red” as eventAand“X”as eventB. Use Formula 6.7 to determine the correct value.
P A B =P A and B P B
P R ed X =P R ed and X
P X = 2
3= 67
The probability of selecting a red ball given an“X”ball has been drawn is .67 or 67%.
For P(Red|Y), first assign “Red” and “Y” to events A and B. Let us define
“Red”as eventAand“Y”as eventB. Use Formula 6.7 to determine the correct value.
P A B =P A and B P B P Red Y =P Red and Y
P Y = 2
7= 29
The probability of selecting a red ball given a“Y”ball has been drawn is .29 or 29%.
ForP(Green|X), first assign“Green”and“X”to eventsAandB. Let us define
“Green” as event A and “X” as event B. Use Formula 6.7 to determine the correct value.
P A B =P A and B P B
P Green X =P Green and X
P X = 1
3= 33
6.8 Conditional Probabilities 181
The probability of selecting a green ball given an“X”ball has been drawn is .33 or 33%.
ForP(Green|Y), first assign“Green”and“Y”to eventsAandB. Let us define
“Green” as event A and “Y” as event B. Use Formula 6.7 to determine the correct value.
P A B =P A and B P B
P Green Y =P Green and Y
P Y = 1
7= 14
The probability of selecting a green ball given a“Y”ball has been drawn is .14 or 14%.■
Before we move on to the final section of this chapter, we need to go back and address the topic of independence again, but now from a more informed vantage point. When we introduced the concept of independence, we stated that two events are independent if the occurrence of one event does not change the likelihood of occurrence for the other event. Determining if two events are independent is often times not a straightforward decision. With a little reflection we may be able to see that selecting a spade and selecting a face card are independent events, but it may be much more difficult to deter- mine merely from what is given in the problem above if a red ball is independ- ent of a ball with an“X”on it. To help matters, it is important to realize that Formula 6.6 is the more generally true formula for determining multiplication problems in probability (the“and”problems). Formula 6.5 is merely a special case version of Formula 6.6. That special case is for situations where events AandBare independent. Look at the two formulas. Here we see that the only difference is that Formula 6.5 refers to P(A), while Formula 6.6 refers to P(A|B). Here we also see that if two events are independent, then the P(A) should equal theP(A|B). That is, if two events are independent, the likelihood that eventAwill occur is the same likelihood if we previously know that event B has or has not occurred. Putting this into the card drawing question, the probability of selecting a spade is 1 out of 4 (25%). This likelihood does not change if we are told ahead of time that the card is a face card; the probability is still 1 out of 4 (or 25%). Therefore, this means we can use the two formulas in combination if we need to determine if two events are independent of each other. We can simply run both formulas and see if they yield the same result. If the two formulas produce the same answer, then we will know that P(A) = P(A|B), and so events A and Bmust be independent of each other.
If, on the other hand, the formulas yield different answers, then we know that
P(A)≠P(A|B), and therefore eventsAandBare dependent upon one another (and so the answer to Formula 6.6 is the correct one). In this way we can use these formulas in combination to help us figure out if two events are independent.
■ Question Given the following probabilistic situation involving balls in an urn, are events Green and X independent? Are events Yellow and Y independent?
15 Green–X
15 Green–Y
10 Red–X
20 Red–Y
25 Yellow–X
15 Yellow–Y
Solution
Step 1.To determine if events“Green”and“X”are independent, we need to first determine the values corresponding to Formulas 6.5 and 6.6. Let us assign
“Green” to be event A and“X” to be eventB. These are P(Green and X), P(Green),P(X),P(Green|X).
P(Green and X) =.15 (given) P(Green) =.3 (Formula 6.1)
P(X) =.5 (Formula 6.1)
P(Green|X) =.15/.5 = .3 (Formula 6.7)
Step 2.Then we need to“run”both formulas. (Recall that we assigned“Green” to be eventAand“X”to be eventB.)
P A and B =P A P B
P Green and X =P Green P X = 3 5 = 15 Formula 6 5 P A and B =P A B P B
P Green and X =P Green X P X = 3 5 = 15 Formula 6 6 Both formulas yield the same answer (.15). This means“Green”and“X”are independent. The likelihood of a“Green”ball occurring is not influenced by the occurrence of a“X”ball (and vice versa).
Now let us apply this same procedure to the second question: Are“Yellow” and“Y”independent?
6.8 Conditional Probabilities 183
Step 1. To determine if events “Yellow” and “Y” are independent, we need to first determine the values corresponding to Formulas 6.5 and 6.6. Let us assign “Yellow” to be event A and “Y” to be event B. These areP(Yellow and Y),P(Yellow),P(Y),P(Yellow|Y).
P(Yellow and Y) =.15 (given) P(Yellow) =.4 (Formula 6.1)
P(Y) =.5 (Formula 6.1)
P(Yellow|Y) =.15/.4 = .375 (Formula 6.7)
Step 2.Then we need to“run”both formulas. (Recall that we assigned“Yellow” to be eventAand“Y”to be eventB.)
P A and B =P A P B
P Yellow and Y =P Yellow P Y = 4 5 = 2 Formula 6 5 P A and B =P A B P B
P Yellow and Y =P Yellow Y P X = 375 5 = 19 Formula 6 6 The formulas do not yield the same answer (.2, .19). The values are close, but technically,“Yellow”and“Y”are not independent. The likelihood of a yellow ball occurring is influenced by the occurrence of a“Y”ball (and vice versa).■