Lesson 8
Conditional Probability
8.1: She Made Some Tarts (5 minutes)
Warmup
The mathematical purpose of this activity is to explore, formally define, and introduce notation for the concept of conditional probability.
Launch
Display the image of a deck of cards.
Explain that there are four suits (spades, hearts, diamonds, clubs) with 13 cards in each suit (ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king) for a total of 52 cards. Hearts and diamonds are always considered red cards while clubs and spades are always considered black cards. Face cards are jack, queen, or king.
Unless otherwise noted, a standard deck of cards in these materials will consist of these 52 cards.
Supports accessibility for: Visualspatial processing; Conceptual processing
Student Facing

Noah will select 1 card at random from a standard deck of cards. Find the probabilities. Explain or show your reasoning.
 \(P(\text{the card is a queen})\)
 \(P(\text{the card is a heart})\)
 \(P(\text{the card is a queen and heart})\)
 Elena pulls out only the hearts from the deck and sets the rest of the cards aside. She shuffles the hearts and draws one card. What is the probability she gets a queen?
Student Response
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Activity Synthesis
The purpose of this discussion is to formally define conditional probability and to introduce notation for conditional probability. Tell students that “conditional probability is defined as the probability of an event happening under the condition that another event occurs.” Use Elena’s situation as an example to show the notation for conditional probability: \(P(\text{the card is a queen  the card is a heart})\) which can be read as “the probability that the card is a queen under the condition that the card is a heart has occurred.” In this situation, Elena made sure the card would be a heart by removing the other suits and drawing only from those cards.
Display \(P(\text{the card is a heart  the card is a queen})\) for all to see and ask students to explain what it means, then ask students to find the probability. (It means the probability that the card is a heart under the condition that it is a queen. It has the value of \(\frac{1}{4}\) since there is one heart card among the 4 queens.) This is the notation we will use throughout this unit to represent conditional probability. Ask students “How would you write what is the probability that a card is an 8 under the condition that is a spade?” (\(P(\text{the card is an 8 the card is a spade})\)).
8.2: Under One Condition (10 minutes)
Activity
The mathematical purpose of this activity is to make sense of conditional probability and the relationship between conditional probabilities and the probability that two events both occur. When students do the same type of calculation several times and observe that \(P(\text{A and B}) = P(\text{A  B}) \boldcdot P(B)\) for events A and B they are engaging in MP8 because they are looking for and expressing regularity in repeated reasoning.
Launch
Remind students about the parts of a standard deck of cards including the working definition of "face card" as a jack, queen, or king of any suit. Tell students that they will be introduced to new notation for probability in the form \(P(\text{A  B})\) and explain that the vertical line inside the parentheses can be read as “under the condition that the next event occurs” or “given that the next event occurs.” For example, \(P(\text{A  B})\) can be read as "the probability of Event A under the condition that Event B occurs." Display the example for all to see.
Supports accessibility for: Language; Organization
Student Facing
Kiran notices that the probabilities from the warmup can be arranged into at least two equations.
\(P(\text{the card is a queen and heart}) = P(\text{the card is a queen  the card is a heart}) \boldcdot P(\text{the card is a heart})\) since \(\frac{1}{52} = \frac{1}{13} \boldcdot \frac{13}{52}\).
\(P(\text{the card is a queen and heart}) = P(\text{the card is a heart  the card is a queen}) \boldcdot P(\text{the card is a queen})\) since \(\frac{1}{52} = \frac{1}{4} \boldcdot \frac{4}{52}\).
Kiran wonders if it is always true that \(P(\text{A and B}) = P(\text{A  B}) \boldcdot P(B)\) for events A and B. He wants to check additional examples from drawing a card from a deck.
 If Event A is “the card is black” and Event B is “the card is a king,” does the equation hold? Explain or show your reasoning.
 If Event A is “the card is a face card” and Event B is “the card is a spade,” does the equation hold? Explain or show your reasoning.
Student Response
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Anticipated Misconceptions
Some students may have difficulty understanding the newlyintroduced notation \(P(\text{A  B})\). Prompt them to think about how they would use the notation to write "the probability of rolling a 6 on a standard number cube under the condition that the number rolled is an even number." Emphasize that the condition of rolling an even number comes after the vertical line, and then show students the correct notation: \(P(\text{6even number})\). In addition, prompt students to record or explain what \(P(\text{evennumber is greater than 3})\) means in words.
Activity Synthesis
The purpose of this discussion is to understand how to calculate conditional probability and how conditional probability is related to independence. Ask students, “Does Kiran's equation always work? Explain your reasoning” (I think it always works because the left and right side of the equation are really two different ways of saying the same thing. If you want to know the probability of event A and event B happening then you can just look at all the probability that event A happens under the condition that event B happens then multiply by the probability of event B happening.). Confirm with students that Kiran’s equation does always work.
Ask
 “A box contains 8 crayons: 3 red, 2 blue, 1 black, 1 green, and 1 yellow. Remove one crayon from the box at random and set it apart, then remove a second crayon from the box. Use Kiran’s equation to find \(P(\text{the second crayon is blue and the first crayon is blue})\). Explain or show your reasoning.” (\(\frac{1}{28}\) since \(P(\text{the second crayon is blue  the first crayon is blue}) = \frac{1}{7}\) and \(P(\text{the first crayon is blue}) = \frac{2}{8}\).)
 “How is the probability of choosing the second crayon dependent on the first crayon?” (It is dependent because the probability that the second crayon is blue could be \(\frac{1}{7}\) or \(\frac{2}{7}\) depending on whether or not the first crayon is blue.)
 “How could you represent the crayon situation using a tree diagram?” (When you list out the probabilities of the first crayon, only 2 of the 8 are blue. For each of the picks there are 7 choices so I know there are 56 total possibilities. For each of the 2 picks that were blue 1 out 7 of the crayons that remain are blue so 2 out 56 is equal to \(\frac{1}{28}\). )
8.3: Coin and Cube (15 minutes)
Activity
The mathematical purpose of this activity is to use conditional probability to investigate independence as it relates to the relationship of \(P(\text{A  B})\) to \(P(\text{A})\).
Launch
Arrange students in groups of 2. Tell students there are many possible answers for the questions. After quiet work time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Follow with a wholeclass discussion.
Design Principle(s): Optimize output (for explanation); Cultivate conversation
Supports accessibility for: Visualspatial processing; Conceptual processing
Student Facing
A coin is flipped, then a standard number cube is rolled. Let A represent the event “the coin lands showing heads” and B represent “the standard number cube lands showing 4.”
 Are events A and B independent or dependent? Explain your reasoning.
 Find the probabilities
 \(P(\text{A})\)
 \(P(\text{B})\)
 \(P(\text{A  B})\)
 \(P(\text{B  A})\)
 Describe the meaning of the events “not A” and “not B” in this situation, then find the probabilities.
 \(P(\text{A  not B})\)
 \(P(\text{B  not A})\)
 Are any of the probabilities the same? Is there a relationship between those situations? Explain your reasoning.
Student Response
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Student Facing
Are you ready for more?
Students are writing programs in their robotics class using two different programming languages that are very similar. 70% of the programs are written in the first programming language and the other 30% percent are written in the second programming language. The newer robots only accept programs written in the first programming language and the older robots only accept programs written in the second programming language.
In this robotics class, Andre’s job is to determine which programming language each program is written in. He determines the programming language correctly 90% of the time for each of the programs. If Andre determines that a program is written in the second programming language, what is the probability that the program is actually written in the first programming language? Explain your reasoning.
Student Response
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Anticipated Misconceptions
Some students may not know what it means for events to be independent or dependent. Refer them to the previous lesson or display the definitions of independent and dependent events for all to see. The term independent events is defined as two events from the same experiment for which the probability of one event is not affected by whether the other event occurs or not, and the term dependent events is defined as two events from the same experiment for which the probability of one event depends on the result of the other event.
Activity Synthesis
The purpose of this discussion is for students to understand the relationship of \(P(\text{A  B})\) to \(P(\text{A})\) for independent events.
Here are some question for discussion.
 “Think of two independent events A and B. What is \(P(\text{A  B})\) and \(P(\text{A})\)?” (I thought about event A as pulling a name out of a bag with ten different names in it and event B as drawing a face card. \(P(\text{A  B})\) is \(\frac{1}{10}\) and \(P(\text{A})\) is also \(\frac{1}{10}\).)
 “If A and B are independent events, what is the relationship of \(P(\text{A  B})\) to \(P(\text{A})\)? Explain your reasoning.” (They are equal since the probability of A happening is not influenced by whether B occurs or not.)
 “When A and B are dependent events, what is the relationship of \(P(\text{A  B})\) to \(P(\text{A})\)? Explain your reasoning” (When the event are dependent, the probabilities are not equal because event B impacts the probability of event A. If event B did not impact the probability of event A then the events would be independent.)
 Look at the values for \(P(\text{A  B})\) and \(P(\text{A  not B})\) that you found in questions 2 and 3. Why are the values equal? (These are equal because the probability of flipping heads is the same regardless of what happened with the number cube.)
 Look at the values for \(P(\text{B  A})\) and \(P(\text{B  not A})\) that you found in questions 2 and 3. Why are the values equal? (These are equal because the probability of getting a 4 is not changed by the coin showing tails.)
Lesson Synthesis
Lesson Synthesis
Display the statement, “Six students (Andre, Claire, Diego, Elena, Han, and Jada) have been selected to participate in a drawing to receive free movie tickets. Two different students will receive free movie tickets.” for all to see.
 “Why isn’t it reasonable to roll a number cube twice to determine the winners?” (It is not reasonable to roll a number cube twice because that does not guarantee that two different names will be chosen.)
 “What is a method that you could use to choose two different names to win the movie tickets?” (You could put the names in a hat and pick one randomly. Then, without putting the first name back in, you could pick a second name.)
 “Using this new method, are the events, picking Andre first and picking Elena second, dependent or independent? Explain your reasoning.” (Dependent, since Elena couldn’t be picked second if her name is picked first, so it depends on what name is picked first.)
 “How would you find the probability of choosing Claire second under the condition that Elena was chosen first?” (The probability of choosing Claire under the condition that Elena was chose first is \(\frac{1}{5}\) because there are only five names left in the hat and only one of them is Elena.)
8.4: Cooldown  Soccer Games (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
A conditional probability is the probability that one event occurs under the condition that another event occurs.
For example, we will remove two marbles from a jar that contains 3 green marbles, 2 blue marbles, 1 white marble, and 1 black marble. We might consider the conditional probability that the second marble we remove is green given that the first marble removed was green. The notation for this probability is \(P(\text{green second  green first})\) where the vertical line can be read as “under the condition that the next event occurs” or “given that the next event occurs.” In this example, \(P(\text{green second  green first}) = \frac{2}{6}\) since we assume the condition that the first marble drawn was green has happened, so the second draw only has 2 possible green marbles left to draw out of 6 marbles still in the jar.
To find the probability of two events happening together, we can use a multiplication rule:
\(P(\text{A and B}) = P(\text{A  B}) \boldcdot P(\text{B})\)
For example, to find the probability that we draw two green marbles from the jar, we could write out the entire sample space and find the probability from that or we could use this rule.
\(P(\text{green second and green first}) = P(\text{green second  green first}) \boldcdot P(\text{green first})\)
Since the probability of getting green on the first draw is \(\frac{3}{7}\) and the conditional probability was considered previously, we can find the probability that both events occur using the multiplication rule.
\(P(\text{green second and green first}) = \frac{2}{6} \boldcdot \frac{3}{7}\)
This tells us that the probability of getting green marbles in both draws is \(\frac{1}{7}\) (since this is equivalent to \(\frac{6}{42}\)).
In cases where events A and B are independent, \(P(\text{A  B}) = P(\text{A})\) since the probability does not change whether B occurs or not. In these cases, the multiplication rule becomes:
\(P(\text{A and B}) = P(\text{A}) \boldcdot P(\text{B})\)
For example, when flipping a coin and rolling a standard number cube, the events “getting a tails for the coin” and “getting 5 for the number cube” are independent. That means we can find the probability of both events occurring to be \(\frac{1}{12}\) by using the multiplication rule.
\(P(\text{heads and 5}) = \frac{1}{2} \boldcdot \frac{1}{6}\)