4. Conditional Probability in the Uniform Probability Model
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Chapter 6
4.
Conditional Probability in the Uniform Probability Model
Conditional probability is a central concept in the world of statistics and plays a crucial role in making decisions based on partial information. Using tools like Venn diagrams, tables, and tree diagrams can simplify the process of understanding and calculating these probabilities. For instance, a Venn diagram visually represents the overlap of different events, while tables and tree diagrams offer structured ways to organize and analyze potential outcomes. Gaining proficiency in these methods allows one to interpret and predict a wide range of scenarios in various fields, from finance to medicine.
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Lesson Settings & Tools
| | 9 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Conditional Probability in the Uniform Probability Model
Slide of 9
In this lesson, a variety of conditional situations will be modeled by using tree diagrams, frequency tables, and Venn diagrams.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
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Detecting if an Email Is Spam
Spam filters determine whether an email is spam by checking it for some words that appear more frequently in spam emails. The following set of information is known.
- 50 % of emails are spam.
- 15 % of spam emails contain the word
free.
- 1 % of non-spam emails contain the word
free.
Jordan gets an email with the word free
in it.
a Draw a tree diagram to visualize the situation.
b What is the probability that the email is spam?
Discussion
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Reviewing the Intuition Behind the Conditional Probability Formula
Recall the formula for the conditional probability.
P(B|A)=P(AandB)/P(A), where P(A) ≠ 0
The intuition behind the formula can be visualized by using Venn diagrams. Consider a sample space S and the events A and B such that P(A)≠0.
Assuming that event A has occurred, the sample space is reduced to A.
It means that the probability that event B can happen is reduced to the outcomes in the intersection of events A and B, or A⋂ B.
The possible outcomes are given by P(A) and the favorable outcomes by P(A⋂ B). Therefore, the conditional probability formula can be obtained using the Probability Formula.
P(B|A)=P(AandB)/P(A)
Example
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Using a Venn Diagram to Find Conditional Probability
Dominika and her friends, 10 people in total, want to play basketball. They decide to form two teams randomly. To do so, each draws a card from a stack of 10 cards numbered from 1 to 10.
- People who draw odd numbers will be on Team Red.
- People who draw even numbers will be on Team Blue.
a Dominika excitedly draws the first card. What is the probability that she draws the number 1? Write the probability as a fraction in its simplest form.
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b If Dominika is on Team Red, what is the probability that she drew the number 1? Write the probability as a fraction in its simplest form.
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Hint
a Determine the sample space.
b Start by drawing a Venn diagram to represent the situation. Use the diagram to determine the sample space of the situation.
Solution
a Since Dominika is the first one to draw a card, there are 10 options for her. Therefore, the sample space for this case consists of numbers from 1 to 10.
S = {1,2,3, ..., 10 } The favorable outcome is to draw the number 1. Using the Probability Formula, the probability of drawing 1 is the ratio of the number of favorable outcomes 1 to the total number of outcomes 10. P(drawing1)= 1/10
b The given situation will be represented by using a Venn diagram. The following two events will be examined.
A: & drawing an odd number B: & drawing the number1 The rest of the possible outcomes will be shown outside of the events A and B.
Since Dominika is on Team Red, the sample space is reduced to the outcomes in A.
There are five odd numbers in the new sample space and only one favorable outcome.
Therefore, using the Probability Formula, the probability that Dominika drew the number 1, given that the number drawn is odd, is found as follows. P(B|A) = 1/5
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Practice Finding Conditional Probability
The applet shows the probabilities of two events in a Venn diagram. Calculate the conditional probabilities. If necessary, round the answer to two decimal places.
Example
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Using a Table to Find Conditional Probabilities
After reading an article about the famous wreck of the Titanic, Paulina concluded that the rescue procedures favored the wealthier first-class passengers. She then finds some data on the survival of the Titanic passengers.
| Survived | Did Not Survive | Total | |
|---|---|---|---|
| First Class Passengers | 201 | 123 | 324 |
| Second Class Passengers | 118 | 166 | 284 |
| Third Class Passengers | 181 | 528 | 709 |
| Total | 500 | 817 | 1317 |
Use this data to investigate the probabilities of surviving the wreck of the Titanic.
a Determine if the events are independent or not. Justify your answer using appropriate probability calculations.
| A: Passenger Survived | |
|---|---|
| B: First Class Passenger | |
| C: Second Class Passenger | |
| D: Third Class Passenger |
b Did all passengers aboard the Titanic have the same probability of surviving? Justify your answer.
Answer
a Table:
| A: Passenger Survived | |
|---|---|
| B: First Class Passenger | Dependent, P(A|B) ≠ P(A) |
| C: Second Class Passenger | Dependent, P(A|C) ≠ P(A) |
| D: Third Class Passenger | Dependent, P(A|D) ≠ P(A) |
b No, see solution.
Hint
a Use the fact that two events A and B are independent if P(A|B)=P(A).
b Compare the conditional probabilities P(A|B), P(A|C), and P(A|D).
Solution
a If two events are independent, the occurrence of one of the events does not affect the occurrence of the other. In other words, the probability that A occurs given that B has already occurred is the same as the probability that A occurs.
| Fraction | Decimal | |
|---|---|---|
| P(A) | 500/1317 | ≈ 0.380 |
| P(A|B) | 201/324 | ≈ 0.620 |
| P(A|C) | 118/284 | ≈ 0.415 |
| P(A|D) | 181/709 | ≈ 0.255 |
Therefore, events B, C, and D each have an effect on event A, meaning that A is a dependent event. In simpler terms, a passenger's chance of surviving depended on what class they were traveling in.
| A: Passenger Survived | |
|---|---|
| B: First Class Passenger | Dependent, P(A|B) ≠ P(A) |
| C: Second Class Passenger | Dependent, P(A|C) ≠ P(A) |
| D: Third Class Passenger | Dependent, P(A|D) ≠ P(A) |
b In the previous part, the conditional probabilities P(A|B), P(A|C), and P(A|D) were found.
P(A|B)≈ 0.620 P(A|C)≈ 0.415 P(A|D)≈ 0.255 Comparing these probabilities, it can be concluded that not all passengers aboard the Titanic had the same chance of surviving. P(A|B) > P(A|C) > P(A|D) 0.620 > 0.415 > 0.255 The first class passengers had the greatest chance of being rescued.
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Practice Finding Conditional Probability
The applet shows the frequency of each event in a table. Calculate the conditional probability asked in the applet. If necessary, round the answer to two decimal places.
Example
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Using a Tree Diagram to Find Conditional Probabilities
Tadeo searches the Internet to check how effective Drug A is compared to Drug B. He finds a research paper about the drugs that gives the following information.
- One third of the participants received Drug A, one third received Drug B, and one third received a placebo.
- 7 % of participants who received Drug A, 11 % of participants who Drug B, and 5 % of participants who received the placebo reported recovery from their condition.
Help Tadeo answer the following questions.
a Draw a tree diagram to represent the situation.
b What is the probability that a participant did not recover if they received Drug B?
c Suppose that the number of participants is 6000. What is the number of participants that received Drug A and did not recover?
Answer
a Example Tree Diagram:
b P(NR | B) = 0.67
c 1580
Hint
a Start by considering the first point of the information Tadeo found. How many outcomes are there initially?
b Determine if it is a conditional probability or not. Use the tree diagram to find it.
c Start by finding the probability of a participant receiving Drug A and not recovering.
Solution
a Consider the first point of the information found by Tadeo.
- One third of the participants received Drug A, one third recieved Drug B, and one third received a placebo.
Since there are three possible outcomes, start by drawing three branches and labeling them. A : & Receiving Drug A B : & Receiving Drug B C : & Receiving the placebo Additionally, since the drugs were distributed among participants evenly, each branch has the same probability, 13.
Each of these outcomes will have two further outcomes — recovered R or not recovered NR. Therefore, two more branches will be drawn for each case.
Each branch should have a probability value on it. Notice that these probabilities are conditional probabilities.
Since there are only two possible outcomes after a certain medicine is received, these outcomes are complements of each other. Therefore, by the Complement Rule, the following equations can be written. P(NR|A) = 1- P(R|A) P(NR|B) = 1- P(R|B) P(NR|C) = 1- P(R|C) Next, these conditional probabilities can be calculated using the information from the second point.
- 7 % of participants who received Drug A, 11 % of participants who Drug B, and 5 % of participants who received the placebo reported recovery from their condition.
P(R|A)=P(RandA)/P(A)
SubstituteII
P(A)= 1/3, P(RandA)= 0.07
P(R|A)=0.07/13
▼
Simplify right-hand side
P(R|A)= 0.21
| Conditional Probability Formula | Substitute | Evaluate | Probability of the Complement | |
|---|---|---|---|---|
| P(R|A) | P(RandA)/P(A) | 0.07/13 | 0.21 | 1-0.21 = 0.79 |
| P(R|B) | P(RandB)/P(B) | 0.11/13 | 0.33 | 1-0.33 = 0.67 |
| P(R|C) | P(RandC)/P(C) | 0.05/13 | 0.15 | 1-0.15 = 0.85 |
Finally, the tree diagram can be completed.
b A participant's well-being varies depending on the drug they received. The probability that a participant did not recover if they received Drug B is a conditional probability because knowing that a participant received Drug B changes the probability of recovery.
P(NR|B) Recall that each final branch of the tree diagram represents a conditional probability. The branch between B and NR will represent the probability of a participant receiving Drug B and not recovering.
Therefore, the conditional probability that a participant did not recover if they received Drug B is 0.67.
c Consider the path through A to NR.
1/3 * 0.79 * 6000
▼
Simplify
1580
Closure
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Detecting if an Email Is Spam
Like Venn diagrams and frequency tables, tree diagrams relate the probability of a conditional event to a subset of the event occurring. With this in mind, reconsider the example given at the beginning of the lesson. These three points about spam emails are known.
- 50 % of emails are spam.
- 15 % of spam emails contain the word
free.
- 1 % of non-spam emails contain the word
free.
Jordan gets an email with the word free
in it.
a Draw a tree diagram to visualize the situation.
b What is the probability that the email is spam? If necessary, round the answer to the two decimal places.
Answer
a Example Tree Diagram:
b P(S|C) ≈ 0.94
Hint
a Start by finding the probabilities that will be written on branches.
b Determine the paths of the tree diagram that lead to emails that contains the word
free.
Solution
a Start by defining the events.
Therefore, P(S), P(C|S), and P(C|NS) are given. P(S) & = 50 % , or 0.5 P(C |S) & = 15 % , or 0.15 P(C|NS) & = 1 % , or 0.01 Notice that NS is the complement of S and NC is the complement of C. Recall that the probability of the complement of an event is 1 minus the probability of the event. Therefore, the probabilities of the complements of the given events can be found. ccc P(S) & & P(NS) [0.4em] 0.5 & ⇔ & 1- 0.5 = 0.5 [0.6em] [-0.4em] P(C |S) & & P(NC |S) [0.4em] 0.15 & ⇔ & 1- 0.15 = 0.85 [0.6em] [-0.4em] P(C|NS) & & P(NC|NS) [0.4em] 0.01 & ⇔ & 1- 0.01 = 0.99
With the events and probabilities determined, a tree diagram can be drawn as shown.
b Jordan gets an email that contains the word
free.The probability that this email is spam needs to be calculated.
P(S | C) = ?
To find it, the tree diagram drawn in the previous part will be used. Consider all the paths that lead to event C, the event that an email contains the word free.
P(S|C) = 0.075/0.08
CalcQuot
Calculate quotient
P(S|C) = 0.9375
RoundDec
Round to 2 decimal place(s)
P(S|C) ≈ 0.94
freehas a 0.94 probability that it is a spam email.
Conditional Probability in the Uniform Probability Model
Exercise 2.1
The following table shows two events. Event A is colored in red and Event B in blue.
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We want to determine the probability of B happening given that A has happened. Before that, we need to define the events.
Defining the Events
As we can see, event A covers all the possible outcomes that contain at least one die showing a 3. Event B covers all outcomes where the sum of the dice is 9. We get the following events. A: &The rolled dice show at least one3. B: &The dice add up to9. To determine P(B|A), we can use the following formula. P(B|A)=P(A and B)/P(A) We can determine the probabilities on the right-hand side using the given diagram.
Determining the Probabilities
Examining the diagram, we can count the number of outcomes for each event. Be careful not to double count the 6.
Now, we can define the probabilities we need. As we can see, 11 out of the 36 total outcomes result in Event A happening. As for Event B, there are four favorable outcomes. However, for both A and B to happen, we only have two favorable outcomes. We get the following probabilities. P(A)& =11/36 [0.7em] P(A and B)& =2/36
Calculating P(B|A)
Finally, we can calculate the probability we are asked.
The probability of P(B|A) is 211.
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Consider the following Venn diagram for the two events A and B.
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The conditional probability P(B|A) can be written as the following formula. P(B|A)=P(A and B)/P(A) This formula divides the probability of A and B by the probability of A. However, we can also find P(B|A) by dividing the number of observations in A and B by the total number of observations in A. P(B|A)=n(A and B)/n(A) Let's visualize the situation.
Since we know that P(B|A)=0.2, we can solve for n(A) in our equation.
As we can see, the number of observations in A, n(A), is 80. This means the observations in A but not B must be 80-16=64. Let's add this to the Venn diagram.
Since 64 corresponds to 50 % of the observations, we know that the total number of observations in the entire sample space, n(S), must be 64* 2=128. Now we can find the number of observations that belong to B but not A by calculating 25 % of the total number of observations.
There are 32 observations that belong to event B but not event A.
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The following diagrams show a study on whether students at a school play an instrument or play sports. However, both diagrams are incomplete.
Use the given information to calculate the following probabilities. Answer with a fraction in its simplest form.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"P(B|A)=","formTextAfter":null,"answer":{"text":["\\dfrac{16}{41}"]}}
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Notice that both the table and the Venn diagram have insufficient information on their own. However, when they are both considered, we can work out the missing information. According to the Venn diagram, 16 students play both an instrument and a sport.
This corresponds to the top left cell of the frequency table, (Yes,Yes).
From the Venn diagram, we also know that 80 students took the survey. This number should equal the sum of the four cells in the table. Since we know the count of three of the cells, we can determine the count in the fourth cell by subtracting them from 80. 80-16-30-9=25 We can now complete both the Venn diagram and the table.
To calculate P(A|B), we need to find P(AandB) and P(B). These values can be found from the diagram. P(AandB) = 16/80 [0.7em] P(B) = 46/80 Finally, we can calculate P(A|B).
In the previous section we identified the missing information in the diagrams. Let's remind ourselves of that.
To calculate P(B|A), we need to find P(AandB) and P(A). These values can be found from the diagram. P(AandB) = 16/80 [0.7em] P(A) = 41/80 Let's substitute these values into the formula for the conditional probability.
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Conditional Probability in the Uniform Probability Model
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