Probability. VCE Maths Methods - Unit 2 - Probability

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Probability Probability Tree diagrams La ice diagrams Venn diagrams Karnough maps Probability tables Union & intersection rules Conditional probability Markov chains 1

Probability Probability is the mathematics of chance. While an outcome of a random event can only be can t determined in advance, we can use a prediction based on experimental or theoretical probability. Probabilities can be expressed as fractions, decimals or percentages. 0 = no chance at all, 100% = absolutely certain. The chances of an event happening or not happening add to make 1. 0% - no chance 25% - unlikely 50% - even chance 75% - likely 0 Pr (E) 1 Pr (E') = 1 - Pr (E) 100% - certain 2

Probability Experimental probability: based on measurement of past outcomes. For example, a 15% chance of a wet grand final day (based on the last 100 years). Theoretical probability: based on the number of favourable outcomes from the total set of possible outcomes. For example, a 1 in a 1000 chance of buying the winning ticket in a raffle. The long term expected value is found by multiplying the probability & number of trials. For example, buying 20 game tickets with a 1 in 10 chance, it is expected that 2 prizes would be won. Experimental probability = Number of favourable outcomes observed Total number of trials Theoretical probability = Number of possible favourable outcomes Total number of possible outcomes Pr (E) = n (E) n (ε) Expected value = E(x) = n Pr(x) 3

Calculating probability - tree diagrams Shows all possible outcomes for sequential events. Each branch on the tree represents a possible outcome. For example, tossing three coins: Each outcome has a 1 in 8 chance (1/2 x 1/2 x 1/2 = 1/8). H T H T H T H T H T H T H T HHH HHT HTH HTT THH THT TTH TTT Pr (3 heads) = 1 8 Pr (2 heads) = 3 1 8 = 3 8 Pr (1 head) = 3 1 8 = 3 8 Pr (0 heads) = 1 8 4

Calculating probability - la ice diagrams Shows the possible outcomes from two sequential events. Easier to draw than a tree diagram if there is a large number of outcomes. For example, rolling two dice & finding the product. 1 2 3 4 5 6 1 1 2 3 4 5 6 2 2 4 6 8 10 12 3 3 6 9 12 15 18 4 4 8 12 16 20 24 The chance of scoring 6: 1 Pr (6) = 4 36 = 1 9 The chance of scoring 36: 5 5 10 15 20 25 30 6 6 12 18 24 30 36 Pr (36) = 1 1 36 = 1 36 5

Venn diagrams A useful visual tool to show the relationship between two sets or events. (Swimming) S S A' n = 2 S A n = 8 (Athletics) A S ' A n = 6 S ' A' n = 9 25 students in the class 10 competed in the swimming sports (Set S) 14 competed in the athletics sports (Set A) 8 students competed in both (Set S A) (S A) = the intersection of events S & A. Two events are mutually exclusive if there is no intersection of the two sets. (If no students did both sports) 6

Union & intersection rules What is the chance that a student that is randomly selected: a) Is a swimmer? b) Is an athlete? c) Is a swimmer and an athlete? d) Is a swimmer or an athlete? Pr(S)= 10 = 40% 25 Pr = 14 = 56% 25 Pr = 8 25 = 32% (Intersection) Pr = 10 25 + 14 25 8 25 = 16 Pr(A B)= Pr(A)+Pr(B) Pr(A B) Pr(A B)=1 Pr(A' B ') 25 = 64% (Union) 7

Karnough maps Used to show the number of elements of two overlapping sets. A A 25 students in the class 10 competed in the swimming sports (Set S) 14 competed in the athletics sports (Set A) 8 students competed in both (Set S A) S 8 2 10 S 6 9 15 14 11 25 8

Karnough maps B B A n(a B) n(a B ') n(a) A n(a' B) n(a' B ') n(a') n(b) n(b ') n 9

Probability tables Used to show the probability of two overlapping events occurring. B B A Pr(A B) Pr(A B ') Pr(A) A Pr(A' B) Pr(A' B ') Pr(A') Pr(B) Pr(B ') 1 10

Conditional probability Two events (A & B) are considered to be independent if the probability of event A occurring has no influence over event B occurring. If the events A & B are not independent, then they are said to be conditional. Pr(A B)= Pr(A B) Pr(B) Pr(A B)= Pr(A B) Pr(B) For example, a student in the class that competed at the athletics - what is the chance that they were also at the swimming sports? Pr(S A)= Independent events: Pr(A B)= Pr(A) Pr(B) Pr(S A) Pr(A) = 8 25 14 25 Conditional events: = 8 = 4 14 7 11

Markov chains W 1 For example, the chance of a team winning is 80% following a win, but only 40% following a loss. What is the chance that after winning the first game, the team also wins the third? This could be expressed as a tree diagram... 80% 20% W 2 W 2 80% 20% 40% 60% W 3 W 3 W 3 W 3 Pr(W 2 W 3 ) = 80% x80% = 64% Pr(W 2 W 3 ) = Pr(W 3 W 2 ) Pr(W 2 ) Pr(W W ') = 80% x20% = 16% 2 3 Pr(W W ') = Pr(W ' W ) Pr(W ) 2 3 3 2 2 Pr(W ' W ) = 20% x40% = 8% 2 3 Pr(W ' W ) = Pr(W W ') Pr(W ') 2 3 3 2 2 Pr(W ' W ') = 20% x60% = 12% 2 3 Pr(W ' W ) = Pr(W W ') Pr(W ') 2 3 3 2 2 Pr(W 3 )= Pr(W 2 W 3 ) + Pr(W 2 ' W 3 ) = 64% + 8% = 72% 12

Markov chains Used to find the long term probability of a sequence of a number of repeated events, where the conditional probabilities remain constant. Pr(B) = Pr(B A) Pr(A)+Pr(B A') Pr(A') Pr(B) Pr(B ') = Pr(B A) Pr(B A') Pr(B ' A) Pr(B ' A') n Pr(A) Pr(A') S n = (Result) State matrix (n states) T= Transition matrix (probabilities) S n =T n S 0 S 0 = State matrix (initial state) 13

Markov chains From the previous example: Winning Losing T= Transition matrix (probabilities) 0.8 0.4 0.2 0.6 2 1 0 S 0 = State matrix (initial state) = 0.72 0.28 S 2 = State (Result) matrix (after 2 more games) As the number of states is increased, the overall probabilities will converge to defined limits. n =1,Pr = 0.800 n = 4,Pr = 0.675 Overall chance of winning n =2,Pr = 0.720 n =15,Pr = 0.667 = 0.4 / (0.4 + 0.2) n = 3,Pr = 0.688 Following a win Following a loss The first game was a win n =100,Pr = 0.667 = 2/3 14