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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.**

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

in it. *free*

a Draw a tree diagram to visualize the situation.

b What is the probability that the email is spam?

Recall the formula for the conditional probability.

$P(B∣A)=P(A)P(AandB) ,whereP(A) =0$

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(AandB) $

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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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.

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.

$P(drawing1)=101 $

b The given situation will be represented by using a Venn diagram. The following two events will be examined.

$A:B: drawing an odd numberdrawing 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.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.

It is practical to model many real-life situations with a uniform probability model, even though they may not actually be uniform after close inspection. Examine the next example and interpret the results.

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.

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.

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).

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.

$P(A),P(A∣B),P(A∣C),andP(A∣D) $

To do so, the Probability Formula will be used. $P=Total number of outcomesNumber of favorable outcomes $

Now, examine the given table to find the probabilities of each event. Adding the event labels to the table can make finding the probabilities a little easier.
Recall that when finding a conditional probability, the sample space is reduced.
Having found the probabilities, note that none of the conditional probabilities is equal to 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) |

b In the previous part, the conditional probabilities P(A∣B), P(A∣C), and P(A∣D) were found.

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.

- 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?

a **Example Tree Diagram:**

b P(NR∣B)=0.67

c 1580

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.

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.

$A:B:C: Receiving Drug AReceiving Drug BReceiving the placebo $

Additionally, since the drugs were distributed among participants evenly, each branch has the same probability, $31 .$
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(R∣A)=P(A)P(RandA) $

SubstituteII

$P(A)=31 $, P(R and A)=0.07

$P(R∣A)=31 0.07 $

Simplify right-hand side

P(R∣A)=0.21

Conditional Probability Formula | Substitute | Evaluate | Probability of the Complement | |
---|---|---|---|---|

P(R∣A) | $P(A)P(RandA) $ | $31 0.07 $ | 0.21 | 1−0.21=0.79 |

P(R∣B) | $P(B)P(RandB) $ | $31 0.11 $ | 0.33 | 1−0.33=0.67 |

P(R∣C) | $P(C)P(RandC) $ | $31 0.05 $ | 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.

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.

$P(AandNR)=31 ⋅0.79 $

The number of study participants who received Drug A and did not recover can be found by multiplying this probability by the total number of participants, 6000 people.
$31 ⋅0.79⋅6000$

Simplify

1580

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

in it. *free*

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.

a **Example Tree Diagram:**

b $P(S∣C)≈0.94$

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.

a Start by defining the events.

$P(S)P(C∣S)P(C∣NS) =50%,or0.5=15%,or0.15=11%,or0.01 $

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. With the events and probabilities determined, a tree diagram can be drawn as shown.

b Jordan gets an email that contains the word

The probability that this email is spam needs to be calculated.free.

The probability of either of these two paths occurring is the sum of the products of the probabilities on the paths.free.

has a 0.94 probability that it is a spam email.free

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