Unit 8, Lesson 9
Using Tables for Conditional Probability
Geometry
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Find the value of each expression mentally.
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:
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 |
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
| 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 |
A blood bank in a region has some information about the blood types of people in its community. Blood is grouped into types 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 that person having each blood type and Rh factor combination is shown in the table.
| O | A | B | AB | total | |
|---|---|---|---|---|---|
| Rh+ | 0.374 | 0.357 | 0.085 | 0.034 | |
| Rh- | 0.066 | 0.063 | 0.015 | 0.006 | |
| total | 1 |
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.
The last two probabilities can also be found using the multiplication rule.
For example, substituting values into