STA 2023
Chapter 3 - Probability
· Events, Sample Spaces, and Probability (3.1)
o Experiment – method of observation that leads to an outcome
o Sample Point – possibility of an experiment
o Sample Space – all possibilities of an experiment
§ S represents sample space
o Venn Diagram – graphical display that contains all possible outcomes
o Example – One suit of a deck of cards
§ S = {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}
§ P(S) = 1 à read “the probability that S occurs is 1”
§ What is the probability that we draw a 3? P(3) =
§ What is the probability that we draw a face card? P(F) =
§ What is the probability that we do not draw a 3 or 4? P(Not 3 or 4) = =
o Properties of a Sample Point
§ 0 £ P(sample point) £ 1 (probability of an outcome is between 0 and 1)
§ P(S) = S P(sample point) = 1 (probability of the sample space is 1)
o Example – Exercise 3.2 – page 117
a. What is this type of diagram called? Venn Diagram
b. Suppose the sample points are equally likely. Find P(A) and P(B). P(A) = P(4) + P(5) + P(6) = .10 + .10 + .10 = .30, while P(B) = P(6) + P(7) = .10 + .10 = .20.
c. Suppose P(1) = P(2) = P(3) = P(4) = P(5) = .05, and P(6) = P(7) = P(8) = P(9) = P(10) = .15. Find P(A) and P(B). P(A) = P(4) + P(5) + P(6) = .05 + .05 + .15 = .25, while P(B) = P(6) + P(7) = .15 + .15 = .30.
· Unions and Intersections (3.2)
o Union – (AÈB) occurs if A occurs, B occurs, or both occur
§ (AÈB) is read “A union B” or “A or B”
o Intersection – (AÇB) occurs if A and B both occur
§ (AÇB) is read “A intersect B” or “A and B”
o Example – Standard deck of playing cards – draw one card
§ What is the probability that we draw a card that is red or black? P(RÈB) = P(card is red or black or both) = 1
§ What is the probability that we draw a card that is red and black? P(RÇB) = P(card is red and black) = 0 (no cards are red AND black)
§ What is the probability that we draw a card that is red and a king? P(RÇK) = P(card is a red king) =
§ What is the probability that we draw a card that is a diamond and a 9? P(DÇ 9) = P(card is a nine of diamonds) =
§ What is the probability that we draw a card that is a diamond or a 9? P(DÈ 9) = P(card is a nine or a diamond or both) =
· Complementary Events (3.3)
o Complement – AC occurs if A does not occur
§ AC is read “A complement”
o Properties of Complementary Events
§ P(A) + P(AC) = 1
§ P(A) = 1 – P(AC) or P(AC) = 1 – P(A)
o Example – Standard Deck of Cards
§ Find P(RC). P(RC) =
§ Find P(DC). P(DC) =
§ Find P(4C). P(4C) =
§ Find P[(4Ç5) C]. P[(4Ç5) C] = 1
§ Find P[(FÇR) C]. P[(FÇR) C] =
§ Find P[(4ÇS) C]. P[(4ÇS) C] =
· The Additive Rule and Mutually Exclusive Events (3.4)
o Additive Rule of Probability – useful for solving “or” probabilities
§ P(AÈB) = P(A) + P(B) – P(AÇB)
o Mutually Exclusive – A and B are mutually exclusive if P(AÇB) = 0
o Probability of Union of Two Mutually Exclusive Events
§ P(AÈB) = P(A) + P(B)
o Example – Offensive Efficiency of a Football Team on a drive, given that P(FG) = .35 and P(TD) = .20.
§ What is the probability that a team scores on an offensive possession? P(Scores) = P(FGÈ TD) = P(FG) + P(TD) – P(FGÇ TD) = .35 + .20 – 0 = .55
· Conditional Probability (3.5)
o Conditional Rule of Probability
§ P(A|B) =
§ P(A|B) is read “conditional probability that even A will occur given that event B has already occurred”
o Example – Kobe Bryant’s field goal shooting statistics
§ Given that Kobe Bryant has scored a field goal, what is the probability that it was a three-pointer? Assume P(FG) = .452, P(2ÇFG) = .406, and P(3ÇFG) = .046. P(3|FG) = = = .102
· The Multiplicative Rule and Independent Events (3.6)
o Multiplicative Rule of Probability – useful for solving “and” probabilities
§ P(AÇB) = P(A) P(B|A) or P(AÇB) = P(B) P(A|B)
o Tree Diagram
o Independent Events
§ If events A and B are independent then P(AÇB) = P(A) P(B)
§ The converse is also true
§ Mutually exclusive events are dependent events (converse not necessarily true)
§ Looking only at a Venn diagram, we can determine whether or not two events are mutually exclusive, however, we cannot determine whether or not two events are independent
o Example – Standard Deck of Cards
§ Are the events of drawing a heart and a 4 independent? P(H) = , P(4) =, so P(H)*P(4) == P(HÇ 4). Yes, they are independent.
§ Are the events of drawing a heart and a 4 mutually exclusive? Since P(HÇ 4) =¹ 0, so they are not mutually exclusive.
§ Are the events of drawing a spade and diamond independent? P(S) = , P(D) =, so P(S)*P(D) =¹ 0 = P(SÇ D), so they are not independent.
§ Are the events of drawing a spade and diamond independent? Since P(SÇ D) = 0, the events are mutually exclusive. Furthermore, since the events are mutually exclusive, we know that the events are also dependent, which agrees with our previous answer.
o Example – Independent Events
§ Suppose that the living members of Pink Floyd (Roger, David, Richard, and Nick) decide to do a reunion tour, and because of their highly volatile personalities, the probability that each member quits the band after any show is as follows: P(Roger quits) = .50, P(David quits) = .40, P(Richard quits) = .20, and P(Nick quits) = .10. Assuming that the band members act independently of one another, find the probability that all members of the band quit immediately after the next show. We are trying to find P(Roger quits Ç David quits Ç Richard quits Ç Nick quits). Since they are independent this is equivalent to P(Roger quits)*P(David quits)*P(Richard quits)*P(Nick quits) = (.50)(.40)(.20)(.10) = .004.
· Random Sampling (3.7)
o Random Sample – when all elements of a population have equal probability of being selected
o Factorial – n! = n(n-1)…1
§ 0! = 1
§ n! is read “n factorial”
o Example – Using factorials
§ Calculate 6!. 6! = 6*5*4*3*2*1 = 720
§ Calculate 10!. 10! = 10*9*8*7*6! = 10*9*8*7*720 = 3,628,800
o Combination – how many ways we can select n objects from a set of N (N > n)
§ =
§ Alternative Notation: or
§ is read “N choose n”
o Example – Basic Combinations
§ How many ways can we choose two marbles from a set of five? Here, N=5 and n=2, so = === 10.
§ How many possible Florida lottery combinations are there? Here, N=53 and n=6, so ==== 22,957,480.
· Some Counting Rules (3.8)
o Permutation – how many ways can we select and order n objects from a set of N
§ =
§ is read “N permute n”
o Example – Executive Board for a campus organization
§ A campus organization consisting of 12 members wishes to select 4 members for the positions of President, Vice President, Secretary, and Treasurer. In how many possible ways can this be accomplished? Here, we have N=12 and n=4, so === 11,880.
o Multiplicative Rule – how many ways we can select one element from each of k sets where ni objects are in set i
§ Number of ways = n1*n2*…*nk
o Example – Eating
§ Suppose you are out to a nice dinner and you are to choose one appetizer (of 3 possible), one main course (of 5 possible), one drink (of 4 possible), and one dessert (of 3 possible). How many possible combinations can be formed? Using the multiplicative rule, the number of combinations that can be formed is 3*5*4*3 = 180.
o Partitions Rule – number of ways to sort all elements of a set of N objects into k groups, where ni objects are in set I
§ Number of ways = where n1 + n2 + … + nk = N
o Example – Offensive Football Team Positions
§ Suppose I have a team of 15 players in which I need to choose 11 of them to start on offense and further choose 1 quarterback, 2 running backs, 3 wide receivers, and 5 linemen. How many different starting lineups are possible? Using the partitions rule, we have = 37,837,800 possible lineups.
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