few_shots: 
Example:
Problem/activity:
You’re mixing ingredients for cookies. The recipe says to combine 6 tablespoons, or ¾ stick, of butter with 1 cup of sugar. You accidentally mix in a whole stick of butter (8 tablespoons) with the cup of sugar. How can you fix this?


Description: 
Recognize and represent proportional relationships between quantities.

Answer: 
No. Usually problems that teach recognizing and representing proportional relationships between quantities involve deciding whether two quantities are in a proportional relationship or identifying the constant of proportionality, and then representing proportional relationships by equations or explaining proportional relationship using coordinate graphs.

Now, assess whether the following problem or activity aligns with the provided description.

shot_info: fl_problem_000972_neg_0

6
standard_description: Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer.
problem_activity: Activity
The mathematical purpose of this activity is to use a two-way table as a sample space to decide if events are independent and to estimate conditional probabilities.
Making spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Give students 10 minutes of quiet time to work the problems and then pause for a brief whole-class discussion.
Action and Expression: Provide Access for Physical Action.
Provide access to tools and assistive technologies. Some students may benefit from a checklist or list of steps to be able to use the spreadsheet technology.
Supports accessibility for: Organization; Conceptual processing; Attention
Student Facing
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.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "row"}}, {"_attributes": {"scope": "col"}, "_value": "no more symptoms"}, {"_attributes": {"scope": "col"}, "_value": "symptoms persist"}, {"_attributes": {"scope": "col"}, "_value": "total"}]}]}], "tbody": [{"tr": [{"th": [{"_attributes": {"scope": "row"}, "_value": "given medicine"}], "td": [{"_value": "31"}, {"_value": "26"}, {"_value": "57"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given placebo"}], "td": [{"_value": "16"}, {"_value": "42"}, {"_value": "58"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "total"}], "td": [{"_value": "47"}, {"_value": "68"}, {"_value": "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.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "row"}}, {"_attributes": {"scope": "col"}, "_value": "no more symptoms"}, {"_attributes": {"scope": "col"}, "_value": "symptoms persist"}, {"_attributes": {"scope": "col"}, "_value": "total"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given medicine"}], "td": [{"_value": "0.27"}, {}, {"_value": "0.50"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given placebo"}], "td": [{}, {}, {}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "total"}], "td": [{}, {}, {"_value": "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.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "row"}}, {"_attributes": {"scope": "col"}, "_value": "no more symptoms"}, {"_attributes": {"scope": "col"}, "_value": "symptoms persist"}, {"_attributes": {"scope": "col"}, "_value": "total"}]}]}], "tbody": [{"tr": [{"th": [{"_attributes": {"scope": "row"}, "_value": "given medicine"}], "td": [{"_value": "0.54"}, {}, {}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given placebo"}], "td": [{}, {}, {"_value": "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.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "row"}}, {"_attributes": {"scope": "col"}, "_value": "no more symptoms"}, {"_attributes": {"scope": "col"}, "_value": "symptoms persist"}, {"_attributes": {"scope": "col"}, "_value": "total"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given medicine"}], "td": [{}, {}, {}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given placebo"}], "td": [{}, {}, {}]}]}]}]}
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.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "row"}}, {"_attributes": {"scope": "col"}, "_value": "no more symptoms"}, {"_attributes": {"scope": "col"}, "_value": "symptoms persist"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given medicine"}], "td": [{"_value": "0.66"}, {}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "given placebo"}], "td": [{}, {}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "total"}], "td": [{}, {"_value": "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?
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Student Facing
Are you ready for more?
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.
{"table": [{"_attributes": {"border": "1"}, "thead": [{"tr": [{"th": [{"_attributes": {"scope": "row"}}, {"_attributes": {"scope": "col"}, "_value": "going to prom"}, {"_attributes": {"scope": "col"}, "_value": "not going to prom"}, {"_attributes": {"scope": "col"}, "_value": "total"}]}]}], "tbody": [{"tr": [{"th": [{"_attributes": {"scope": "row"}, "_value": "grade 11"}], "td": [{}, {}, {"_value": "127"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "grade 12"}], "td": [{}, {}, {"_value": "116"}]}, {"th": [{"_attributes": {"scope": "row"}, "_value": "total"}], "td": [{"_value": "124"}, {"_value": "119"}, {"_value": "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?
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
Students might struggle with understanding the notation for conditional probability. Tell these students that
\(P(\text{A | B})\)
means the probability of A happening under the condition that B happens.
Activity Synthesis
The purpose of this discussion is for students to interpret probabilities in a two-way table.
Ask students to interpret each of the probabilities they found in questions 1, 2, and 4. For example, “What does the probability in question 1a mean in the context of the problem?”
Here are sample interpretations of the probabilities for questions 1, 2, and 4.
The probability that a test subject chosen at random from all of the subjects still had symptoms after the test.
The probability that a test subject chosen at random from all of the subjects was among those given medicine and still had symptoms after the test.
The probability that a test subject chosen at random from all of the subjects still had symptoms after the test or was given a placebo. 1
The probability that a test subject chosen at random from those who were given medicine had symptoms after the test.
The probability that a test subject chosen at random from those who were given placebo had no more symptoms after the test
The probability that a test subject chosen at random from those who had symptoms after the test was given medicine.
The probability that a test subject chosen at random from those who had no more symptoms after the test was given placebo.
Display what Jada did when creating the top left cell in table for all to see:
\(\frac{P(\text{no more symptoms and given medicine})}{P(\text{given medicine})}\)
Here are some questions for discussion.
“What do you notice about what Jada did and
\(P(\text{no more symptoms | given medicine})\)
?” (I noticed that they are equal.)
“What is an equation that represents this situation using A and B for the events?” (
\(P(\text{A | B}) = \frac{P(\text{A and B})}{P(\text{B})}\)
)
"What is the relationship between this equation and the multiplication rule?" (From the multiplication rule, we know that
\(P(\text{A and B}) = P(\text{A | B}) \boldcdot P(\text{B})\)
. If we divide both sides by
\(P(\text{B})\)
, we get the equation that Jada used.)
“How did you determine whether the events were dependent or independent?” (I determined that they were dependent events because
\(P(\text{symptoms persist}) = 0.6\)
and
\(P(\text{symptoms persist | given medicine}) = 0.46\)
were not equal. If the events were independent, these probabilities would be equal.)
Speaking: MLR8 Discussion Supports.
Use this routine to support whole-class discussion. For each observation that is shared, ask students to restate what they heard using precise mathematical language. Consider providing students time to restate what they hear to a partner before selecting one or two students to share with the class. Ask the original speaker if their peer was accurately able to restate their thinking. Call students’ attention to any words or phrases that helped to clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.
Design Principle(s): Support sense-making


aligns: no