Lesson 9
Using Tables for Conditional Probability
 Let’s use tables to estimate conditional probabilities.
9.1: Math Talk: Fractions in Fractions
Evaluate each expression mentally.
\(\frac{7}{11} \div \frac{8}{11}\)
\(\frac{\left( \frac{1}{3} \right)}{\left( \frac{2}{3} \right)}\)
\(\frac{\left( \frac{1}{4} \right)}{\left( \frac{1}{2} \right)}\)
\(\frac{\left( \frac{3}{8} \right) \left( \frac{8}{19} \right)}{\left( \frac{5}{19} \right)}\)
9.2: A Possible Cure
A pharmaceutical company is testing a new medicine for a disease using 115 test subjects. Some of the test subjects are given the new medicine and others are given a placebo. The results of their tests are summarized in the table.
no more symptoms  symptoms persist  total  

given medicine  31  26  57 
given placebo  16  42  58 
total  47  68  115 
 Divide the value in each cell by the total number of test subjects to find each probability to two decimal places. Some of the values have been completed for you.
no more symptoms symptoms persist total given medicine 0.27 0.50 given placebo total 1 If one of these test subjects is selected at random, find each probability:
 \(P(\text{symptoms persist})\)
 \(P(\text{given medicine and symptoms persist})\)
 \(P(\text{given placebo or symptoms persist})\)

From the original table, divide each cell by the total for the row to find the probabilities with row conditions. Some of the values have been completed for you.
no more symptoms symptoms persist total given medicine 0.54 given placebo 1  \(P(\text{symptoms persist  given medicine})\)
 \(P(\text{no more symptoms  given placebo})\)
 Jada didn’t read the instructions for the previous problem well and used the table she created on the first problem to divide each cell by the probability total for each row. For example, in the top left cell she calculated \(0.27 \div 0.5\). Complete the table using Jada’s method.
no more symptoms symptoms persist total given medicine given placebo What do you notice about this table?

From the original table, divide each cell by the total for the column to find the probabilities with column conditions. Some of the values have been completed for you.
no more symptoms symptoms persist given medicine 0.66 given placebo total 1  \(P(\text{given medicine  symptoms persist})\)
 \(P(\text{given placebo  no more symptoms})\)

Are the events “symptoms persist” and “given medicine” independent events? Explain or show your reasoning.

Based on your work, does being given this medicine have an impact on whether symptoms persist or not?
Consider the data collected from all students in grades 11 and 12 and their intentions of going to prom.

65% of the students who are going to prom are from grade 12. Complete the table.
going to prom not going to prom total grade 11 127 grade 12 116 total 124 119 243 
Based on your results, what is the probability that a student from this group, selected at random, is a student in grade 11 that did not go to prom?
9.3: The Blood Bank
A blood bank in a region has some information about the blood types of people in its community. Blood types are grouped into type O, A, B, and AB. Each blood type either has the Rh factor (Rh+) or not (Rh). If a person is randomly selected from the community, the probability of their having each blood type and Rh factor combination is shown in the table.
O  A  B  AB  

Rh+  0.374  0.357  0.085  0.034 
Rh  0.066  0.063  0.015  0.006 
 What does the 0.085 in the table represent?
 Use the table or create additional tables to find the probabilities, then describe the meaning of the event.
 \(P(\text{O})\)
 \(P(\text{Rh+})\)
 \(P(\text{O and Rh+})\)
 \(P(\text{O or Rh+})\)
 \(P(\text{O  Rh+})\)
 \(P(\text{Rh+  O})\)
Summary
Organizing data in tables is a useful way to see information and compute probabilities. Consider the data collected from all students in grades 11 and 12 and their intentions of going to prom.
going to prom  not going to prom  total  

grade 11  43  84  127 
grade 12  81  35  116 
total  124  119  243 
A student is randomly selected from this group of students. We can find a number of probabilities based on the table.
\(P(\text{grade 12 and going to prom}) = \frac{81}{243}\) since there are 81 students who are both in 12th grade and going to prom out of all 243 students in the group who could be selected.
\(P(\text{grade 12}) = \frac{116}{243}\) since there are 116 grade 12 students out of the entire group.
\(P(\text{going to prom}) = \frac{124}{243}\) since there are 124 students going to prom out of the entire group.
\(P(\text{grade 12 or going to prom}) = \frac{159}{243}\) since there are 159 students in grade 12 or going to prom (43 from grade 11 going to prom, 81 from grade 12 going to prom, and 35 from grade 12 not going to prom).
\(P(\text{going to prom  grade 12}) = \frac{81}{116}\) represents the probability that the chosen student is going to prom under the condition that they are in grade 12. The group we are considering is different for this probability since the condition is that they are in grade 12. Imagine all grade 12 students are in a room and we are only selecting from this group. Therefore 81 students are going to prom from this group out of 116 students in the group.
\(P(\text{grade 11  going to prom}) = \frac{43}{124}\) represents the probability that the chosen student is in grade 11 under the condition that they are going to prom. In this case, we imagine that everyone going to prom is gathered in a room and we are selecting one student from this group. The probability that this student is in grade 11 when just considering this group is \(\frac{43}{124}\).
The last two probabilities can also be found using the multiplication rule.
\(P(\text{A and B}) = P(\text{A  B}) \boldcdot P(\text{B})\) can be rewritten \(P(\text{A  B}) = \frac{P(\text{A and B})}{P(\text{B})}\).
For example, substituting values into \(P(\text{going to prom  grade 12}) = \frac{P(\text{going to prom and grade 12})}{P(\text{grade 12})}\) gives \(\frac{81}{116} = \frac{\left(\frac{81}{243}\right)}{\left(\frac{116}{243}\right)}\) which is a true equation.
Glossary Entries
 conditional probability
The probability that one event occurs under the condition that another event occurs.
 dependent events
Dependent events are two events from the same experiment for which the probability of one event depends on whether the other event happens.
 independent events
Independent events are two events from the same experiment for which the probability of one event is not affected by whether the other event occurs or not.