5/22

Do Now 5/22 Are the events “Favorite book is The Hidden Gnome” and “Favorite movie is Henry Porter” independent? Essential Question: What is the Addition Rule of probability?

Agenda Do Now Good Things! Recap: Notes: Independent Events Notes: Addition Rule Conditional Probability Practice Problems + Review as whole group

GOOD THINGS!!

Recap: Independent Events Events A and B are independent events if: What does this mean????? The probability of one event happening given that another event happened is the exact same as if the other one never happened!!

Independent Events A school cafeteria serves breakfast to high school students and then to middle school students. One morning the cafeteria served breakfast to 40 high school students and then 60 middle school students. Of those 100 breakfasts, 50 were bacon and egg biscuits – 20 of those were served to high school students and 30 were served to middle school students. Consider the following events: B: A bacon and egg biscuit is served. MS: A middle school student is served. p(A|B) = p(A), independent if A is “bacon and eggs” and b is “middle school” p(bacon eggs | middle school) = 30/60 and p(bacon eggs) = 50/100 = ½ for both (independent!) opposite is also true p(middle school | bacon eggs) = 30/50 and p(middle school) = 60/100  both = .6, so they are independent Are these events independent???

Addition Rule P(A or B) = P(A) + P(B) – P(A and B) We have been finding the probability of two events being independent We can use the Addition Rule to determine the probability of event A or B happening P(A or B) = P(A) + P(B) – P(A and B) How many probabilities are there TOTAL involved in this equation? 4! If we know 3 of them, we can figure out the 4th. Why do we subtract the last part? Think about a Venn Diagram

Addition Rule P(A or B) = P(A) + P(B) – P(A and B) The addition rule has FOUR parts P(A or B) = P(A) + P(B) – P(A and B) If we have 3 of the 4 parts, we can solve for the missing part! Key words: OR and AND Remember, if 2 events are independent we can solve for the “and” part by multiplying

Group Practice Students at Rolling Hills High School receive an achievement award for either performing community service or making the honor roll. The school has 500 students and 180 of them received the award. There were 125 students who performed community service and 75 students who made the honor roll. What is the probability that a randomly chosen student at Rolling Hills High School performed community service and made the honor roll?

Group Practice P(A or B) = P(A) + P(B) – P(A and B) 1. Define the sample space  500 students total 2. Identify the probabilities A = “performed community service.” B = “made the honor roll.” Event “A or B” is “performed community service or made the honor roll” 180 people got the award, so this must be 180/500 P(A or B) = P(A) + P(B) – P(A and B)

Group Practice P(A or B) = P(A) + P(B) – P(A and B) Substitute the known probabilities.

Group Practice The probability that a randomly chosen student at Rolling Hills High School has performed community service and is on the honor roll is

Addition Rule Practice South Point High School has 147 tenth-grade students. There are 85 tenth graders in physical education, 91 tenth graders in homecoming parade, and 130 tenth graders in physical education or homecoming parade. What is the probability that a randomly chosen tenth grader at South Point High School is in physical education and homecoming parade? (130/147) = (85/147) + (91/147) - X P(A or B) = P(A) + P(B) – P(A and B)

Addition Rule – Independent Events P(A or B) = P(A) + P(B) – P(A and B) Since they are independent, we can MULTIPLY them to find the probability of both A AND B.

Conditional Probability The chances of something happening given that something else already happened Also called “dependent” probability P(B|A) = P(A and B) / P(A) We use this when we don’t know the probability of B given A

Conditional Probability A survey revealed that 56% of the students at North Valley High School are involved in a sports team. The survey also showed that 24% of the students at the school are involved in a sports team and also participated in the Prom dance. What is the probability that a student who is involved in a sports team also participated in the Prom dance? P(B|A) = P(A and B) / P(A) P( Prom | sports ) = Probability of BOTH / Probability of just Sports .24 / .56 = .4285 = 42.9%

Practice Problems! www.problem-attic.com/test/zvzft4iq ~30 minutes to complete 7 problems We will review them together at the end of class www.problem-attic.com/test/zvzft4iq