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| 9 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
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.
free.
free.
Jordan gets an email with the word free
in it.
Recall the formula for the conditional probability.
P(B∣A)=P(A)P(A and B), 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(A)P(A and B)
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.
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.
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.
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: Passenger Survived | |
---|---|
B: First Class Passenger | |
C: Second Class Passenger | |
D: Third Class Passenger |
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) |
Fraction | Decimal | |
---|---|---|
P(A) | 1317500 | ≈0.380 |
P(A∣B) | 324201 | ≈0.620 |
P(A∣C) | 284118 | ≈0.415 |
P(A∣D) | 709181 | ≈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) |
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.
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.
Help Tadeo answer the following questions.
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.
P(A)=31, P(R and A)=0.07
Conditional Probability Formula | Substitute | Evaluate | Probability of the Complement | |
---|---|---|---|---|
P(R∣A) | P(A)P(R and A) | 310.07 | 0.21 | 1−0.21=0.79 |
P(R∣B) | P(B)P(R and B) | 310.11 | 0.33 | 1−0.33=0.67 |
P(R∣C) | P(C)P(R and C) | 310.05 | 0.15 | 1−0.15=0.85 |
Finally, the tree diagram can be completed.
Therefore, the conditional probability that a participant did not recover if they received Drug B is 0.67.
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.
free.
free.
Jordan gets an email with the word free
in it.
free.
With the events and probabilities determined, a tree diagram can be drawn as shown.
free.The probability that this email is spam needs to be calculated.
free.
Calculate quotient
Round to 2 decimal place(s)
freehas a 0.94 probability that it is a spam email.
The following table shows two events. Event A is colored in red and Event B in blue.
We want to determine the probability of B happening given that A has happened. Before that, we need to define 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.
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
Finally, we can calculate the probability we are asked.
The probability of P(B|A) is 211.
Consider the following Venn diagram for the two events A and B.
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.
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.
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.