Multiplication Rules for Probability and Independent Events P(event A and event B) 11/15/2018

Independent Events Important question to ask yourself: This means: “Are these two events Independent?” This means: “If one of these events happens, does it affect the probability of the other event happening?” “Or are they Dependent events?” 11/15/2018

Independent/Dependent Examples Experiment: Rolling a single die, two trials Are these events Independent? Event #1: Rolling an odd number on the first roll Event #2: Rolling an even number on the second Event #2: Rolling a “high” number (4, 5, or 6) on the second roll. 11/15/2018

Independent/Dependent Examples Experiment: Pick one card, then pick a second card Are these events Independent or Dependent? Event #1: Your first card is an 8, then you replace the card and the deck is reshuffled. Event #2: Your second card is a face card Event #1: Your first card is an 8 and you hang onto it 11/15/2018

Independent/Dependent Examples Experiment: Draw two Scrabble tiles and (this is a key phrase) without replacement Are these events independent? Event #1: You get an “A” on the first pick Event #2: You get a “B” on the second pick Are these events Independent? Event #1: You get one of the blank tiles on the first pick Event #2: And the other blank tile on the second. 11/15/2018

Multiplication Rule to find 𝑃(𝐴 𝑎𝑛𝑑 𝐵) If the events are Independent events… If the events are Not independent events… 𝑃 𝐴 𝑎𝑛𝑑 𝐵 =𝑃 𝐴 ∙𝑃(𝐵) The probability of event A Times the probability of event B 𝑃 𝐴 𝑎𝑛𝑑 𝐵 =𝑃 𝐴 ∙𝑃 𝐵|𝐴 The probability of event A Times the probability of event B, given that event A has already occurred. The Vertical bar in 𝐵|𝐴 is means “assuming that” 11/15/2018

Example: Rolling a single die twice P(odd and then a 4) P(odd and odd again) Independent events P(odd) is ______ P(4) is ______ P(odd and 4) = P(odd) ∙ P(4) = _____ ∙ _____ = _____ Independent events P(odd) is _____ P(odd and odd) = P(odd) ∙ P(odd) = _____ ∙ _____ = ______ Extend it: P(odd) ten times in a row: ______________ 11/15/2018

Example: Pick two cards With and Without Replacement Draw an ace, replace it in the deck, then draw a ten Draw an ace, keep it, then draw a ten With Replacement Independent Events P(ace) = _____ P(ten) = _____ P(ace and ten) = P(ace) ∙ P(ten) = ______ ∙ ______ = ______ Without Replacement Dependent Events P(ace) = _____ P(ten | ace) = _____ P(ace and ten) = P(ace) ∙ P(ten | ace) = _____ ∙ ______ = _____ 11/15/2018

Example: How many passwords are possible? If repetition allowed or not An 8-character password made of A-Z, 0-9 if repetition allowed If you aren’t allowed to repeat any characters “Repetition allowed” is another way of saying “with replacement” 36 8 What does 2.821109907E12 mean? 28,211,099,070,000, approximately. Same idea as “Without Replacement” 36∙35∙34∙33∙32∙31∙30∙29=30,260,340 The possibilities with repetition allowed are about ____ times as many. 11/15/2018

Example: How many passwords are possible? If repetition allowed or not An 8-character password made of A-Z, 0-9 if repetition allowed If you aren’t allowed to repeat any characters 36 8 possible passwords What is the probability that a randomly generated password will begin with a digit? What is the probability that a randomly generated password will be all digits? 30,260,340 What is the probability that a randomly generated password will begin with a digit? What is the probability that a randomly generated password will be all digits? 11/15/2018

Word Problems Sometimes P(A and B) problems are presented in narrative form. Example: Fair Deal Frank’s car lot has 3 new trucks, 4 used trucks, 5 new cars, and 6 used cars. Choose two keys at random Find P(two new trucks) Find P(two used vehicles) 11/15/2018

Fair Deal Frank, continued P(two new trucks) P(two used vehicles) Dependent events P(new truck) = _______ P(new truck|new truck) = ______ P(two new trucks) = P(new truck) ∙ P(new truck | new truck) = ________ ∙ ________ = ________ Dependent events P(used) = _______ P(used | used) = ______ P(two used vehicles) = P(used)+P(used | used) = ______ ∙ ______ = ______ 11/15/2018

In a large population With 52 cards or a dozen keys or 100 Scrabble® tiles, we have to pay attention to “with replacement” or “without replacement” But in a large population, like choosing three people at random, we don’t worry about it as much. 11/15/2018

Example 4-26: Survey on Stress from Bluman 5th Ed. © McGraw Hill A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week. We are taking three people at random, without replacement. But because the population is so large, we may treat the three selections as independent events. The calculation is a lot simpler this way. Bluman, Chapter 4

Example 4-26: Survey on Stress from Bluman 5th Ed. © McGraw Hill A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week. Bluman, Chapter 4

Problems given as Data Tables Example: a gathering of high income people who signed up for a webinar on how to shelter income from the tax man Men Women Total Doctor 300 200 Lawyer 600 Neither 400 500 11/15/2018

High Income Earners Choose one person at random… P(you picked a woman doctor) Shortcut Not Independent events P(woman) = _______ P(doctor | woman) = ______ P(A and B) = P(A) ∙ P(B | A) = ____ ∙ ____ = _____ Go back to the basic definition of probability: 𝑃 𝑒𝑣𝑒𝑛𝑡 = 𝑚𝑦 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑡𝑜𝑡𝑎𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 “My outcomes” is count of women doctors, right there in a cell. “Total outcomes” in bottom right corner of the chart. 11/15/2018

Complementary Events example Each widget that rolls off the assembly line has a 99% chance of passing inspection. If five widgets are selected at random, what is the probability that at least one of them fails? Are the events “Widget #1 is good”, “#2 is good”, etc. Independent or Dependent? 11/15/2018

Complementary Events in the widget inspection department P(this one is good) = _____ P(all are good, all five out of five) = ____ ∙ ____ ∙ ____ ∙ ____ ∙ ____ = ( _____ )___ = _____ P(at least one is bad) means “Not all are good”. “not all are good” is complement of “all good” P(not all are good) = 1 - P(all are good) = 1 - ____ = _____ 11/15/2018

Same as the widget problem with complementary events “A single die is rolled 10 times. Find the probability of rolling at least one ‘6’.” Complementary Event is ___________________. “If the probability of twins is 5% and there are ten expectant moms in the maternity ward, find P(at least one set of twins).” Complementary event is ___________________. 11/15/2018

Complementary Events example Suppose that a railroad crossing lights and bells and bars have a 99.9% chance of operating successfully each time a train passes by. If there are 5 trains a day x 365 days a year, what is the probability of no malfunctions? P(perfect operation) = (0.99)what exponent? What is the probability of at least one error? It’s the complement of the “perfect operation”. 11/15/2018

Conditional Probability Suppose we know P(A) and we know P(A and B) and want to find the unknown P(B | A). Someone took 𝑃 𝐴 ∙𝑃 𝐵 𝐴 =𝑃 𝐴 𝑎𝑛𝑑 𝐵 And divided each side by 𝑃 𝐴 Getting this result: 𝑃 𝐵 𝐴 = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) 𝑃(𝐴) 11/15/2018

4.3 Conditional Probability Conditional probability is the probability that the second event B occurs given that the first event A has occurred. 23 Bluman, Chapter 4

Example 4-33: Parking Tickets Bluman 5th ed. © McGraw Hill The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket. Bluman, Chapter 4

Bluman’s Example 4-33: Parking Tickets WRITE OUT DEFINITIONS! Let T = gets a ticket event, Let N = parks in a no-parking zone event. What the problem says What it means in algebra Parking in a no-parking zone Call it event N Getting a ticket Call it event T P(no parking zone and ticket is 0.06) P(N and T) = 0.06 P(he has to park in a no-parking zone) is 0.20 P(N) = 0.20 If he parks in a no-parking zone, find probability he will get a ticket P(T | N) = ? Bluman, Chapter 4

Example 4-33: Parking Tickets Bluman 5th ed. © McGraw Hill What the problem says What it means in algebra Parking in a no-parking zone Call it event N Getting a ticket Call it event T P(no parking zone and ticket is 0.06 P(N and T) = 0.06 P(he has to park in a no-parking zone) is 0.20 P(N) = 0.20 If he parks in a no-parking zone, find probability he will get a ticket P(T | N) = ? Bluman, Chapter 4

Example 4-34: Women in the Military A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown. Bluman, Chapter 4

Example 4-34: Women in the Military a. Find the probability that the respondent answered yes (Y), given that the respondent was a female (F). Bluman, Chapter 4

Example 4-34: Women in the Military When you have a data table problem like this, the shortcut is to have Event A determine one row or one column, and compute directly based on that one row or one column. P(Y | F) = Look at Female row only: 8 Yes / 50 Total Bluman, Chapter 4

Example 4-34: Women in the Military b. Find the probability that the respondent was a male (M), given that the respondent answered no (N). Shortcut Shown In red boxes Bluman, Chapter 4