Conditional Probability and Independence Section 5.3 Reference Text: The Practice of Statistics, Fourth Edition. Starnes, Yates, Moore Lesson 6.1.1

Objectives Conditional Probability 2. Independent Events “what's the probability of Event B given that Event A has happened” 2. Independent Events If the chances of event B occurring is not affected by whether event A occurs, then A and B are independent! 3. Tree Diagrams! Map out my probability calculations! General Multiplication Rule P(A ∩ B) = P(A) * P(B I A) Special case of independent events, multiplication rule becomes P(A ∩ B) = P(A) * P(B) 4. Conditional Probability formula

Two way Tables Students in college stats class wanted to find out how common it is for young adults to have their ears pierced. They recorded data on two variables- gender and whether the student had a pierced ear – for all 178 people in class. The two way table below displays the data. Pierced ears? Gender Yes No Total Male 19 71 90 Female 84 4 88 103 75 178

A= male B= pierced ears P(A) = P(male) = P(B) = P(pierced ears) = P(A ∩ B) = P(male and pierced ears) = P(A U B) = P(male or pierced ears) = Now lets turn out attention to some other interesting probability questions…

Conditional Probability The probability that one event happens given that another event is already known to have happened is called Conditional Probability. Suppose we known that event A has happened. Then the probability that event B happens given that event A has happened is denoted by P(B I A) Read | as “given that” or “under the condition that”

Heads Up You might be thinking to yourself in the previous slide and this next slide, “hey, we’ve done this before! Is there a connection between conditional probability and the conditional distributions of Chapter 1?!” Of course! We have been doing probability and conditional probability from the start!

Conditional Probability Conditional probability narrows your focus in onto a specific event. Given a certain condition…. A= male B= pierced ears We can ask… P(A I B) then switch the order… P( B I A)

Conditional Probability and Independence Example: Grade Distributions Consider the two-way table on page 314. Define events E: the grade comes from an EPS course, and L: the grade is lower than a B. Conditional Probability and Independence Total 3392 2952 3656 10000 Total 6300 1600 2100 Find P(L) Find P(E | L) Find P(L | E) P(L) = 3656 / 10000 = 0.3656 P(E | L) = 800 / 3656 = 0.2188 P(L| E) = 800 / 1600 = 0.5000

Conditional Probability and Independence When knowledge that one event has happened does not change the likelihood that another event will happen, we say the two events are independent. Conditional Probability and Independence Definition: Two events A and B are independent if the occurrence of one event has no effect on the chance that the other event will happen. In other words, events A and B are independent if P(A | B) = P(A) and P(B | A) = P(B). Are the events “male” and “left-handed” independent? Justify your answer. Example: P(left-handed | male) = 3/23 = 0.13 P(left-handed) = 7/50 = 0.14 These probabilities are not equal, therefore the events “male” and “left-handed” are not independent.

Independence: Think of it this way In order to know if two events are independent we need to compute probabilities. Not Independent P(left-handed | male) = 3/23 = 0.13 P(left-handed) = 7/50 = 0.14 .𝟏𝟑≠.𝟏𝟒, 𝒕𝒉𝒖𝒔 𝒏𝒐𝒕 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 Independent P(A I B) = 19/100 = .19 P(A) = 19/100 = .19 .𝟏𝟗= .𝟏𝟗, 𝒕𝒉𝒖𝒔 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕

Checking for Independence In order to know if two events are independent we need to compute probabilities. Example: Is there a relationship between gender and handedness? To find out we used CensusAtSchool’s Random Data Selector to chose an SRS of 50 Australian high school students who completed a survey. The two way table displays data on the gender and dominant hand of each student. Dominant Hand Gender Right Left Total Male 20 3 23 Female 4 27 43 7 50

Are the events “male” and “left handed” independent? Suppose we are told that the chosen student is male. From the two way table P(Left handed I male) = 3/23 =0.13 The unconditional probability P(left handed) = 7/50 =0.14 These two probabilities are close, but they’re not equal. So the events “male” and “left handed” are not independent.

Conditional Probability and Independence Tree Diagrams We learned how to describe the sample space S of a chance process in Section 5.2. Another way to model chance behavior that involves a sequence of outcomes is to construct a tree diagram. Conditional Probability and Independence Consider flipping a coin twice. What is the probability of getting heads and heads again? The word “and” implies multiplication! Sample Space: HH HT TH TT So, P(two heads) = P(HH) = 1/4

Conditional Probability and Independence General Multiplication Rule The idea of multiplying along the branches in a tree diagram leads to a general method for finding the probability P(A ∩ B) that two events happen together. Conditional Probability and Independence The probability that events A and B both occur can be found using the general multiplication rule P(A ∩ B) = P(A) • P(B | A) where P(B | A) is the conditional probability that event B occurs given that event A has already occurred. General Multiplication Rule

General Multiplication Rule When we calculate the probability of flipping two heads, we are following the general multiplication rule! P(A ∩ B) = P(A) * P(B I A) Read as: “The Probability of A intersect B equals the probability of flipping a heads multiplied by the probability of flipping a second heads given that the first flip was heads.”

Teens with online profiles The Pew Internet and American Life Project finds that 93% of teenagers (ages 12 to 17) use the internet, and that 55% of online teens have posted a profile on a social-networking site Question: What percent of teens are online and have posted a profile? Create a tree diagram. Lets do this on the board!

Conditional Probability and Independence Example: Teens with Online Profiles The Pew Internet and American Life Project finds that 93% of teenagers (ages 12 to 17) use the Internet, and that 55% of online teens have posted a profile on a social-networking site. What percent of teens are online and have posted a profile? Conditional Probability and Independence 51.15% of teens are online and have posted a profile.

Check for Understanding A computer company makes desktops and laptop computers at factories in three states- California, Texas, and New York. The California factory produces 40% of the company’s computers, the Texas factory makes 25%, and the remaining 35% are manufactured in New York. Of the computers made in California, 75% are laptops. Of those made in Texas and New York, 70% and 50% respectively, are laptops. All computers are first shipped to a distribution center in Missouri before being sent out to stores. Suppose we select a computer at random from the distribution center. Construct a tree diagram to represent the situation. Find the probability that it’s a laptop from California. Show your work.

Independence: A Special Multiplication Rule Note that this rule only applies to independent events. Lets look at an example. Definition: Multiplication rule for independent events If A and B are independent events, then the probability that A and B both occur is P(A ∩ B) = P(A) • P(B)

The Challenger Disaster On January 28, 1986, Space Shuttle Challenger exploded on takeoff. All seven crew members were killed. Following the disaster, scientists and statisticians helped analyze what went wrong. They determined that the failure of O-ring joints in the shuttle’s booster rockets was to blame. Under cold conditions that day, experts estimated that the probability that an individual O-ring joint would function properly was 0.977. But there were six of these O-ring joints, and all six had to function properly for the shuttle to launch safely. Assuming that O-ring joints succeed or fail independently, What is the probability that the shuttle would launch safely under similar conditions? Example:

The Challenger Disaster Assuming that O-ring joints succeed or fail independently, what is the probability that the shuttle would launch safely under similar conditions? P(joint 1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK) = (0.977) (0.977) (0.977) (0.977) (0.977) (0.977)= 0.87 There is an 87% chance that the shuttle would launch safely under similar conditions (and a 13% chance that it wouldn’t).

Conditional Probability and Independence Calculating Conditional Probabilities If we rearrange the terms in the general multiplication rule, we can get a formula for the conditional probability P(B | A). Conditional Probability and Independence General Multiplication Rule P(A ∩ B) = P(A) • P(B | A) P(A ∩ B) P(A) P(B | A) To find the conditional probability P(B | A), use the formula Conditional Probability Formula =

Conditional Probability and Independence Example: Who Reads the Newspaper? In Section 5.2, we noted that residents of a large apartment complex can be classified based on the events A: reads USA Today and B: reads the New York Times. The Venn Diagram below describes the residents. What is the probability that a randomly selected resident who reads USA Today also reads the New York Times? Conditional Probability and Independence There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.

You Try! We classified U.S. households according to the type of phones they used: Question: What is the probability that a randomly selected household with a landline also has a cell phone? Solution: 𝑃( 𝑐𝑒𝑙𝑙 𝑝ℎ𝑜𝑛𝑒 𝑙𝑎𝑛𝑑𝑙𝑖𝑛𝑒 = 𝑃(𝑙𝑎𝑛𝑑𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑐𝑒𝑙𝑙 𝑝ℎ𝑜𝑛𝑒) 𝑙𝑎𝑛𝑑𝑙𝑖𝑛𝑒 Cell Phone No Cell Phone Total Landline 0.60 0.18 0.78 No Landline 0.20 0.02 0.22 0.80 1.00

Objectives Conditional Probability 2. Independent Events “what's the probability of Event B given that Event A has happened” 2. Independent Events If the chances of event B occurring is not affected by whether event A occurs, then A and B are independent! 3. Tree Diagrams! Map out my probability calculations! General Multiplication Rule P(A ∩ B) = P(A) * P(B I A) Special case of independent events, multiplication rule becomes P(A ∩ B) = P(A) * P(B) 4. Conditional Probability formula

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