Chapter- 3 Statistics Probability topics
- 1. Statistics Chapter 3 : Probability Topics
- 2. Terminology + Probability deals with the chance of an event occurring. + Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment. + A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes. + An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P(A).
- 3. + " ∪ " Event: The Union An outcome is in the event A ∪ B if the outcome is in A or is in B or is in both A and B. For example, let A={1,2,3,4, 5} and B = {4, 5, 6, 7, 8}. A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice. + " ∩ " Event: The Intersection An outcome is in the event A ∩ B if the outcome is in both A and B at the same time. For example, let A and B be {1,2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A ∩ B = {4, 5}. + The complement of event A is denoted A′ (read "A prime"). A′ consists of all outcomes that are NOT in A. Notice that P(A) +P(A′) =1.For example, let S={1,2,3,4,5,6} and let A={1,2,3,4}. Then, A′={5,6}. P(A)= 4 6 , P(A′)= 2 6 ,and P(A) + P(A′) = 4 6 + 2 6 = 1 + The conditional probability of A given B is written P(A | B). P(A | B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A | B) is P(A | B) = 𝑃 𝐴 ∩ 𝐵 𝑃 𝐵 where P(B) is greater than zero. + Suppose we toss one fair, six-sided die. The sample space S = 1, 2 , 3 , 4, 5 , 6 . Let A = face is 2 or 3 and B = face is even ( 2, 4 ,6 ). To calculate P(AIB), we count the number of outcomes 2 or 3 in the sample space B= 2, 4, 6 . Then we divide that by the number of outcomes B( rather than S). P( AIB) = 1/6 3/6 = 1/3 Exercise 1. The sample space S is the whole numbers starting at one and less than 20. S = _____________________________ Let event A = the even numbers and event B = numbers greater than 13. A = _____________________, B = _____________________ P(A) = _____________, P(B) = ________________ A ∩ B = ____________________, A OR B = ________________ P(A ∩ B) = _________, P(A ∪ B) = _____________ A′ = _____________, P(A′) = _____________ P(A) + P(A′) = ____________ P(A | B) = ___________, P(B | A) = _____________; are the probabilities equal?
- 4. Exercise 2. Describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left- handed. + Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities: P(M) = P(F) = P(R) = P(L) = P(M ∩ R)= P(F ∩ L) = P(M ∪ F) = P(M ∪ R) = P(F ∪ L) = P(M’) = P(R|M) = P(F | L) = P(L | F) =
- 5. Independent and Mutually Exclusive Events + Two events are independent if one of the following are true: + P(A|B) = P(A) + P(B|A) = P(B) + P(A ∩ B) = P(A)P(B) Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. + A and B are mutually exclusive events if they cannot occur at the same time. Said another way, If A occurred then B cannot occur and vise-a-versa. This means that A and B do not share any outcomes and P(A ∩ B) = 0 .
- 6. Example Let event G = taking a math class. Let event H = taking a science class. Then, G ∩ H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5 , and P(G ∩ H) = 0.3. Are G and H independent? If G and H are independent, then you must show ONE of the following: + P(G|H) = P(G) + P(H|G) = P(H) + P(G ∩ H) = P(G)P(H) a. Show that P(G|H) = P(G) . b. Show P(G ∩ H) = P(G)P(H) . Solution : P(G|H) = P(G ∩ H) / P( H) = 0.3/0.5 = 0.6 P(G ∩ H) = 0.6 * 0.5 = 0.3 G and H are independent events.
- 7. Exercise 3. Let event C = taking an English class. Let event D = taking a speech class. Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C ∩ D) = 0.225. Justify your answers to the following questions numerically. a. Are C and D independent? b. Are C and D mutually exclusive? c. What is P(D|C) ?
- 8. Two Basic Rules of Probability The Multiplication Rule + If A and B are two events defined on a sample space, then: P(A ∩ B) = P(B)P(A|B) . We can think of the intersection symbol as substituting for the word "and". This rule may also be written as: P(A|B) = P(A ∩ B)/P(B) The Addition Rule + If A and B are defined on a sample space, then: P(A∪B)=P(A)+P(B)- P(A∩B). We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B. + If A and B are mutually exclusive, then P(A ∩ B) = 0. Then P(A ∪ B) = P(A)+P(B)- P(A ∩ B) becomes P(A ∪ B) = P(A)+P(B).
- 9. Example A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty- seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. + a. What is the probability that the member is a novice swimmer? 28 150 + b. What is the probability that the member practices four times a week? + 80 150 + c. What is the probability that the member is an advanced swimmer and practices four times a week? + 40 150 + d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? P(advanced ∩ intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. + e. Are being a novice swimmer and practicing four times a week independent events? Why or why not? No, these are not independent events. P(novice ∩ practices four times per week) = 0.0667 P(novice)P(practices four times per week) = 0.0996 0.0667 ≠ 0.0996
- 10. Exercise 4. Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class | that she enrolls in speech class is 0.25. Let: M = math class, S = speech class, M | S = math given speech + What is the probability that Felicity enrolls in math and speech? Find P(M ∩ S) = P(M | S)P(S). + What is the probability that Felicity enrolls in math or speech classes? Find P(M ∪ S) = P(M) + P(S) - P(M ∩ S). + Are M and S independent? Is P(M | S) = P(M)? + Are M and S mutually exclusive? Is P(M ∩ S) = 0?
- 11. Contingency Tables and Probability Trees + A contingency table provides a way of portraying data that can facilitate calculating probabilities. Example : Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data: The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755. a. Find P(Driver is a cell phone user). 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑒𝑙𝑙 𝑝ℎ𝑜𝑛𝑒 𝑢𝑠𝑒𝑟𝑠 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑠𝑡𝑢𝑑𝑦 = 305 755 b. Find P(Driver had no violation in the last year). 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 ℎ𝑎𝑑 𝑛𝑜 𝑣𝑖𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑠𝑡𝑢𝑑𝑦 = 685 755 c. Find P(Driver had no violation in the last year ∩ was a cell phone user). 280 755 d. Find P(Driver is a cell phone user ∪ driver had no violation in the last year). 305 755 + 685 755 – 280 755 = 710 755 e. Find P(Driver is a cell phone user | driver had a violation in the last year). 25 70 f. Find P(Driver had no violation last year | driver was not a cell phone user) 405
- 12. Exercise 5. Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S. TOTAL each column and each row. Total data = 4,520.7 + Find P(2009 ∩ Robbery) . + Find P(2010 ∩ Burglary). + Find P(2010 ∪ Burglary) . + Find P(2011 | Rape). + Find P(Vehicle | 2008).
- 13. Venn Diagram + A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. P( A) = P( B) = P( A ∩ B) = P( A’) = P ( A ∪ B) =
- 14. Exercise 1. Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time. • the probability that the student belongs to a club. • the probability that the student works part time. • the probability that the student belongs to a club AND works part time. • the probability that the student belongs to a club given that the student works part time. • the probability that the student belongs to a club OR works part time.
- 15. Tree Diagrams + A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
- 16. EXAMPLE + In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows. • Using the tree diagram, calculate P(RR). 𝑃 𝑅𝑅 = 3 11 3 11 = 9 121 • Using the tree diagram, calculate P(RB ∪ BR) . 𝑃 𝑅𝐵 ∪ 𝐵𝑅 = 3 11 8 11 + 8 11 3 11 = 48 121 • Using the tree diagram, calculate P(R on 1st draw ∩ B on 2nd draw) . 𝑃 𝑅 𝑜𝑛 1𝑠𝑡 𝑑𝑟𝑎𝑤 ∩ 𝐵 𝑜𝑛 2𝑛𝑑 𝑑𝑟𝑎𝑤 = 3 11 8 11 = 24 121 • Using the tree diagram, calculate P(R on 2nd draw | B on 1st draw). 𝑃(𝑅 𝑜𝑛 2𝑛𝑑 𝑑𝑟𝑎𝑤 | 𝐵 𝑜𝑛 1𝑠𝑡 𝑑𝑟𝑎𝑤) = 𝑃(𝑅 𝑜𝑛 2𝑛𝑑 | 𝐵 𝑜𝑛 1𝑠𝑡) = 24/88 = 3/11 This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). • Using the tree diagram, calculate P(BB). P(BB) = 64/121 • Using the tree diagram, calculate P(B on the 2nd draw | R on the first draw). P(B on 2nd draw | R on 1st draw) = 8 /11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB).
- 17. Exercise 2. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled. + Suppose that you randomly draw two cards, one at a time, with replacement. + Let G1 = first card is green + Let G2 = second card is green + a. Draw a tree diagram of the situation. + b. Find P(G1 AND G2). + c. Find P(at least one green). + d. Find P(G2|G1). + e. Are G2 and G1 independent events? Explain why or why not.
- 18. Exercise 3. An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. + a. P(RR) = ________ + b. P(RB ∪ BR) = + c. P(R on 2nd | B on 1st) = + d. P(R on 1st ∩ B on 2nd) + e. P( B on 1st ∩ R on 2nd) + f. P(BB)
- 19. Exercise 4. In a management program at Mount Enterprises, 80 % of the trainee are female and 20% male. 90% of the female attended college, and 78% of the male attended college. a. Construct a tree diagram showing all the probabilities. b. Do the joint probabilities total 1?
- 20. Exercise 5. What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart? Construct Venn diagram to portray these outcomes.