4.5 through 4.9 Probability continued…

Today’s Agenda Go over page 158 (49 – 52, 54 – 58 even) Go over 4.5 and 4.6 notes Class work: page 158 (53 – 57 odd). Hw: Review worksheet for Quiz tomorrow.

4.5 Mutually Exclusive Events Mutually Exclusive Events- events that can not occur together. Do not have ANY common outcome. Example: Rolling a die. Let, A = even #, then A = {2, 4, 6} B= odd #, then B = {1, 3, 5} C = less than 5, then C = {1, 2, 3, 4} Which events are mutually exclusive and Why? Which events are not and Why?

Another Example Consider the following two events for a randomly selected adult: Y = this adult has shopped on the Internet at least once N = this adult has NEVER shopped on the Internet. Are the events Y and N Mutually Exclusive? YES!!! Y and N have No common outcomes.

4.6 Independent Versus Dependent Events Two events are said to be Independent if the occurrence of one does NOT affect the Probability of the other. Example: Rolling a die and tossing a coin. Definition: 2 events, A and B, are Independent if either: P(A | B) = P(A) or P(B | A) = P(B) If one above is true, then the other one is true. They mean that the Conditional Prob.= Marginal Prob.

Example of Independence FavorAgainst M1545 F436 Are events F and A independent? According to the definition they will be independent if P(Female) = P(Female| A). Is this true? NO! P(Female) = 40 / 100=.4 and P( Female | A) = 36/81=.444. These are NOT equal and therefore, the two are DEPENDENT EVENTS! Also P(A) does NOT equal P(A | F).

Dependence If the probability of one event Affects the probability of the other, then the 2 events are said to be dependent. In our example, being Female and Agreeing were dependent on each other because if your were Male, the percentages for in Favor or Against were different.

Mutually Exclusive and Dependence Relationship Mutually Exclusive Events (events that have NO common outcomes) are ALWAYS dependent on Each other! Ex 1: Say you roll a di once. A = even #, then A = {2, 4, 6} and B= odd #, then B = {1, 3, 5} are Both Mutually Exclusive events. They are also Dependent because: P(A) does not equal P(A|B). 3/6 does not equal 0.

Dependent or Independent? Ex 2: Imagine you are drawing ONE card from a deck. K = the event of pulling a K F= the event of pulling a 5 Are the two events independent of each other? In other words doesP(K) = P(K | F)? P(K) = 4/52 does not equal P(K | F)= 0/52. So they are Dependent! ( We knew this already since K and F are mutually exclusive).

Dependent or Independent events? Ex 3: Imagine you toss a coin then roll a di. Let A = the event of rolling a 3 on the di and H= the event of having tossed a Head. There are 2* 6 = 12 outcomes. H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6. P(A) = 2/12 = 1/6 and the P(A|H)= 1/6 also! Thus rolling a di and tossing a coin are Independent of each other.

Class work Do page 158 (53 – 57 odd including parts on mutually exclusive and independence) Then do 4.1 to 4.4 worksheet to review for quiz tomorrow. Thanks

4.7 Complementary Events Two mutually exclusive events that taken together include ALL the outcomes for an experiment are called Complementary events. The complement of A is denoted by (pronounced A bar). Two Complementary Events are Always Mutually Exclusive (have NO common outcomes).

The Complement of A, (A bar), contains all the outcomes for an experiment that are not in A. P(A) + P( ) = 1 In other words, P( ) = 1 – P(A). Complementary Events Definitions continued…

Venn Diagram of Complements

Examples Let A= rolling a 2. What is the P(A)? What is the P(A bar)? P(A) = 1/6 P ( A bar) = 5/6 Let B = picking an Ace from a deck. What is P(B)? What is the P(B bar)? P(B) = 4/52 P(B bar) = 48/52.

Another Example Let’s say we have n = 5000 adults of them favor gun control, 1200 of them are against, and 300 have no opinion. Let A = the event that an adult is in Favor. a.) What is A bar? Adult against or no opinion. b.) What are P(A) and P(A bar)? P(A) = 3500/5000 = 70% P(A bar) = =.30 or 30%

Homework Page 158 (47, 48, 59-61, 63, 64)

4.8 Joint Probability (AND) Joint Probability- the Probability of A AND B (both events) occurring. It is the INTERSECTION of two events occurring. The intersection of A and B is ALL outcomes that are Common to BOTH A and B. Notation: A and B

Example Say A = the event of a family owning a DVD player and B= the event of owning a Camera. What is the P(A and B)=? Venn Diagram:

Multiplication Rule The probability of the intersection of 2 events A and B is found by the formula: P(A and B) = P(A) * P(B|A)

Example 1 What is the Probability that if you select one person, the employee is Female AND a college Graduate? P(F and G)=? P(F and G) = P(F) * P(G|F)= 13/40 * 4/13= 4/40 =.10 = 10% GraduateNot GraduateTOTAL Male72027 Female4913 TOTAL112940

Example 1 cont… P(M and G)=? P(M and G) = P(M) * P(G|M)= 27/40 * 7/27= 7/40 =.175= 17.5 % P(F and N)=? P(F and N) = P(F) * P(N|F)= 13/40 * 9/13= 9/40 =.225 = 22.5% NOTE: you don’t need Formula! How could you have gotten these answers from table otherwise? GraduateNot GraduateTOTAL Male72027 Female4913 TOTAL112940

You can make a tree diagram to find ALL Joint Probabilities! GraduateNot GraduateTOTAL Male72027 Female4913 TOTAL112940

Example 2 (no table given) Say we have a box with 20 DVDS (4 of which are Defective). If 2 DVDs are chosen (w/o replacement), what is the probability that BOTH are Defective? Let’s say: G 1 = first DVD is good G 2 = second DVD is good D 1 = first DVD is defective D 2 = second DVD is defective

Example 2 cont… We are being asked to find P(D 1 and D 2 )? Use formula: P(D 1 and D 2 )=P(D 1 ) *P(D 2 |D 1 ) Well, P(D 1 )=4/20. What is P(D 2 |D 1 )? This is a little harder to think of…What is the Probability that the 2 nd DVD is defective given that the 1 st DVD was defective? P(D 2 |D 1 )=3/19! Thus, P(D 1 and D 2 )= 4/20 *3/19 =.0316 = 3.2 %

Conditional Probability P(A|B) We know how to find Conditional Probability when given a chart of data. But if we don’t have chart to find Conditional Probabilities, but we have other info, this formula may help: P(A|B) = P(A and B)/ P(B)

Example Conditional Probability(no chart) The probability that a randomly selected student from a college is a senior is.20. The joint probability that the student is a computer science major AND a senior is.03. Find the conditional probability that the student selected is a computer science major given he/she is a senior. Since no chart, and we have other info in word problem use: P(A|B) = P(A and B)/ P(B) P(Computer Science | Senior) =.03 /.20 =.15 = 15%

Joint Probability of Independent Events Recall Definition: Independent Events means that the occurrence of the 1 st event does NOT affect the probability of the occurrence of the other. Recall Formulas: Joint Probability- P(A and B) = P(A) * P(B|A) Independent Events- P(A | B) = P(A) or P(B | A) = P(B) So if we wanted P(A and B) for two Independent Events… P(A and B)=?

If 2 events are Independent… You can simplify Joint probability equation to the following: P(A and B)= P(A) * P(B) For INDEPENDENT events ONLY! Ex: The Probability that a patient is allergic to penicillin is.20. Say we choose 3 patients. Find P( all 3 are allergic)=? Find P(at least 1 is NOT allergic)=?

SOLUTION Let A = event 1 st patient is allergic B = event 2 nd patient is allergic C = event 3 rd patient is allergic P( all 3 are allergic)= P(A and B and C) Choosing 3 patients is Independent, therefore, P(A and B and C) =.20 *.20 *.20 =.008 =.8% P(at least 1 is NOT allergic)= ? This is a little harder… At least one Not allergic means that we can have 1, 2, or 3 NOT. In other words, we can have 2, 1, or 0 Allergic.

TREE Diagram to get ALL Joint Probabilities.

So…back to our question. P(at least 1 is NOT allergic)= ? P(at least 1 NOT allergic) = P(1, 2, or 3 NOT allergic) = P(everything but 0 people NOT allergic) = 1 – P(A and B and C) = =.992 = 99.2%

Joint Probability of Mutually Exclusive Events Recall Definition: Mutually Exclusive Events have NO common outcomes. So for Mutually Exclusive Events ONLY: P(A and B) = 0 Ex: P(Male and Female) = 0

Homework p. 169 – 171 (76, 78, 80, 82, 86, 90)

4.9 Union of Events and the Addition Rule (OR) The Union of events A and B include ALL outcomes that are either in A OR B or In both A and B. Basically it includes EVERYTHING! Denoted by A U B. Venn Diagram:

Formula: (Addition Rule) P(A or B) = P(A) + P(B) – P(A and B) We have to subtract out intersection since it is counted twice.

Example 1 Let A = person who is a Faculty member and B = person who is in Favor. Given the following data, find the Probability of one person selected being a faculty member OR in favor. Use formula: P(A or B) = P(A) + P(B) – P(A and B) P(A or B)= 70/ /300 – 45/ 300 = =.533 = 53.3 % chance FavorOpposedNeutralTOTAL Faculty Students TOTAL

Putting it all together… P(favor OR student)= ? P(opposed OR neutral)= ? P(neutral OR faculty)=? P(neutral AND faculty)=? P(opposed AND favor)=? P(student | opposed)= ? P(student OR faculty)=? FavorOpposedNeutralTOTAL Faculty Students TOTAL

Another Example of Union Example: A senior citizen center has 300 members. Of them, 140 are Male, 210 take at least one medicine on a permanent basis, and 95 are male AND take at least one medicine on a permanent basis. Let M = a senior who is Male Let F = a senior who is Female Let A = a senior who takes at least one Medicine Let B = a senior who does NOT take any Medicine

Example cont… Describe the union of events M and A. In other words, Find P(M U A). P(M U A) = P(M) + P(A) – P(M and A) P(M U A) = 140/ /300 – 95/300 = 255/300 =.85 = 85 %

Union and Mutually Exclusive Events Mutually Exclusive Events – have NO common outcomes. Union- means “OR”. In other words, you take EVERYTHING. So, if two events are mutually exclusive: P(A or B) = ?

If A and B are Mutually Exclusive. P(A or B) = P(A) + P(B) This is because there is no intersection to subtract out.

Example for Union of Mutually Exclusive Events Probability of a person in favor of genetic engineering is.55. Two people are selected and it is observed whether they are for or against. Note: Being For or Against are mutually exclusive! Draw a tree diagram: P(at least one person favors genetic engineering)= P(AF or FA or FF) = P(AF) + P(FA) + P(FF) = =.7975 = %

Homework p. 178 (106 – 122 even)