Consider a pair of events, A and B. Suppose that we are concerned about the probability of A, given that B has occurred. This problem can be approached using the concept of conditional probability. The basic idea is that the probability of any event occurring often depends on whether or not other events have occurred. For example, a manufacturer planning to introduce a new brand may test-market the product in a few selected stores.
This manufacturer will be much more confident about the brand’s success in the wider market if it is well accepted in the test market than if it is not. The firm’s assessment of the probability of high sales will, therefore, be conditioned by the test-market outcome.
If I knew that interest rates would fall over the next year, I would be far more bullish about the stock market than if I believed they would rise. What I know, or believe, about interest rates conditions my probability assessment of the course of stock prices. Next, we give a formal statement of conditional probability that can be used to determine the effect of prior results on a probability.
Solution Let A be the event “customer wants text messaging” and B be the event
“customer wants photo capability.” Thus,
P1A2 = 0.75 P1B2 = 0.80 and P1A > B2 = 0.65 The required probability is as follows:
P1A < B2 = P1A2 + P1B2 - P1A > B2 = 0.75 + 0.80 - 0.65 = 0.90
Note that the first step was to write the probabilities in mathematical form; then the solution followed directly using Equation 3.8.
Conditional Probability
Let A and B be two events. The conditional probability of event A, given that event B has occurred, is denoted by the symbol P1A u B2 and is found to be as follows:
P1A u B2 = P1A > B2
P1B2 provided that P1B2 7 0 (3.9) Similarly,
P1B u A2 = P1A > B2
P1A2 provided that P1A2 7 0
We can better understand these results and those that follow by considering Table 3.3.
The conditional probability, P1A u B2, is the ratio of the joint probability, P1A > B2, di-vided by the probability of the conditional variable, P1B2. This conditional probability could be thought of as using only the first row of the table that deals only with condition B. A similar analysis could be made for the conditional probability P1B u A2.
Table 3.3 Joint Probability of A and B
A A
B P1A > B2 P1A > B2 P(B) B P1A > B2 P1A > B2 P1B2
P(A) P1A2 1.0
Relative frequencies can also help us understand conditional probability. Suppose that we repeat a random experiment n times, with nB occurrences of event B and nA>B occur-rences of A and B together. Then the proportion of times that A occurs, when B has occurred,
114 Chapter 3 Elements of Chance: Probability Methods
is nA>B>nB, and one can think of the conditional probability of A, given B, as the limit of this proportion as the number of replications of the experiment becomes infinitely large:
nA>B
nB = nA>B>n nB>n
As n becomes large, the numerator and denominator of the right-hand side of this expression approach P1A > B2 and P1B2, respectively.
Example 3.15 Product Choice: Cell Phone Features (Conditional Probability)
In Example 3.14 we noted that 75% of the customers want text messaging, 80% want photo capability, and 65% want both. What are the probabilities that a person who wants text messaging also wants photo capability and that a person who wants photo capability also wants text messaging?
Solution Designating A as text messaging and B as photo capability, we know that P1A2 = 0.75, P1B2 = 0.80, and P1A > B2 = 0.65. The probability that a person who wants photo capability also wants text messaging is the conditional probability of event A, given event B is:
P1A u B2 = P1A > B2
P1B2 = 0.65
0.80 = 0.8125
In the same way, the probability that a person who wants text messaging also wants photo capability is as follows:
P1B u A2 = P1A > B2
P1A2 = 0.65
0.75 = 0.8667
These calculations can also be developed using Table 3.4.
Note that the conditional probability that a person wanting photo capability also wants text messaging is the joint probability 0.65 divided by the probability of a person wanting photo capability, 0.80. A similar calculation can be made for the other conditional probability. We have found that some people believe that using a table such as Table 3.4 provides better motivation and success for solving conditional probability and related problems that follow. Using the table correctly will provide exactly the same results as using the equations. So, if this helps you with these problems you can feel perfectly com-fortable with using tables to solve the problems.
Table 3.4 Joint Probability for Example 3.15
Text Messaging No Text Messaging
Photo 0.65 0.15 0.80
No Photo 0.10 0.10 0.20
0.75 0.25 1.0
The Multiplication Rule of Probabilities
Let A and B be two events. Using the multiplication rule of probabilities, the probability of their intersection can be derived from conditional probability as
P1A > B2 = P1A u B2 P1B2 (3.10) and also as
P1A > B2 = P1B u A2 P1A2
3.3 Probability Rules 115 In the following example we see an interesting application of the multiplication rule of probabilities. We also tie together some ideas introduced previously.
Example 3.17 Sensitive Questions (Multiplication Rule)
Suppose that a survey was carried out in New York, and each respondent was faced with the following two questions:
a. Is the last digit of your Social Security number odd?
b. Have you ever lied on an employment application?
The second question is, of course, quite sensitive, and for various reasons we might expect that a number of people would not answer the question honestly, especially if their response were yes. To overcome this potential bias, respondents were asked to flip a coin and then to answer question (a) if the result was “head” and answer (b) oth-erwise. A yes response was given by 37% of all respondents. What is the probability that a respondent who was answering the sensitive question, (b), replied yes?
Solution We define the following events:
A: Respondent answers yes.
E1: Respondent answers question (a).
E2: Respondent answers question (b).
From the problem discussion we know that P1A2 = 0.37. We also know that the choice of question was determined by a flip of a coin and that P1E12 = 0.50 and P1E22 = 0.50. In addition, we know the answers to question (a). Since half of all Social Security numbers have an odd last digit, it must be that the probability of a yes answer, given that question (a) has been answered, is 0.50—that is, P1A u E12 = 0.50.
However, we require P1A u E22, the conditional probability of a yes response, given that question (b) was answered. We can obtain this probability by using two results from previous sections. We know that E1 and E2 are mutually exclusive and collectively exhaustive. We also know that intersections E1> A and E2> A are mutually exclusive and that their union is A. It therefore follows that the sum of the probabilities of these two intersections is the probability of A, so
P1A2 = P1E1> A2 + P1E2> A2 Next, we use the multiplication rule to obtain
P1E1> A2 = P1A u E12P1E12 = 10.502 10.502 = 0.25 and
P1E2> A2 = P1A2 - P1E1> A2 = 0.37 - 0.25 = 0.12
Example 3.16 Cell Phone Features (Multiplication Rule)
When the conditional probability of text messaging, given photo capability, P1A u B2 = 0.65
0.80 = 0.8125
is multiplied by the probability of photo capability, we have the joint probability of both messaging and photo capability:
P1A > B2 = 10.8125210.802 = 0.65
116 Chapter 3 Elements of Chance: Probability Methods