Minds on! If you choose an answer to this question at random, what is the probability you will be correct? A) 25% B) 50% C) 100% D) 25%

Steps to Solving Probability Problems G = Given: List knowns, draw diagram / table / list R = Required to find: Identify the unknown, look for keywords A = Analyze: Write the formula S = Solve: Plug knowns into formula and compute P = Present: State your answer using a sentence

Warm up #1 In 2016-17, the Toronto Maple Leafs won 19 of their first 50 games. Their leading goal scorer, Leo Komarov, scored at least one goal in 10 of those. Given that the Leafs won their next game, what is the probability that Komarov scored a goal?

Solution Let W be the event that the Leafs won. Let K be the event that Leo Komarov scored. The probability that Komarov scored, given that the Leafs won is 0.53.

Warm up #2 Let H1 be the event that the first card is a heart Find the probability that two hearts are dealt consecutively from a standard deck (no replacement). Let H1 be the event that the first card is a heart Let H2 be the event that the second card is a heart P(H1 ∩ H2) = P(H2 | H1) x P(H1) = 12/51 x 13/52 = 156/2652 = 0.06

4.5 Finding Probability Using Tree Diagrams and Outcome Tables Learning goal: Calculate probabilities using tree diagrams and outcome tables Questions? pp. 235 – 238 #1, 2, 4, 6, 7, 9, 10, 19 MSIP / Home Learning: pp. 245 – 249 #2, 3, 5, 7, 9, 12, 13a, 14

Tree Diagram 1 if you flip a coin twice, you can model the possible outcomes using a tree diagram or an outcome table resulting in 4 possible outcomes Flip 1 Flip 2 Simple Event H HH T HT TH TT H T T Toss 1 Toss 2

How to Draw a Tree Diagram Each branch represents an event Draw the branch for the first event Add a branch to each for the second event Repeat for additional events Follow each branch to list the outcomes

Tree Diagram 2 if you roll 1 die and then flip a coin there are 12 possible outcomes H T (1,H) 1 (1,T) H T (2,H) 2 (2,T) H T (3,H) 3 (3,T) (4,H) H T 4 (4,T) H T (5,H) 5 (5,T) (6,H) H T 6 (6,T)

Tree Diagram 2 (cont’d) The sample space is all ordered pairs in the form (d,c), where d represents the roll of a die c represents the flip of a coin e.g., (4,T) represents rolling a 4 then flipping a tail There are 6 x 2 = 12 possible outcomes What is P(odd roll, head) = ? There are 3 possible outcomes for an odd die and a head: (1, H), (3, H) and (5, H) So the probability is 3/12 or ¼

Multiplicative Principle for Counting The total number of outcomes is the product of the number of possible outcomes at each step in the sequence If a is selected from A, and b selected from B… n(a,b) = n(A) x n(B) This assumes that each outcome has no influence on the next outcome

Examples Pt 1 How many outcomes are there when rolling 2 dice? How many sets of initials are there? 26 x 26 = 676 What if letters cannot be repeated? This adds a restriction on the 2nd letter 26 x 25 = 650

Examples Pt 2 Airport codes are arrangements of 3 letters e.g., YOW=Ottawa, YYZ=Toronto How many are possible? 26 choices for each of the 3 positions 26 x 26 x 26 = 17 576 How many with no repeated letters? 26 x 25 x 24 = 15 600

Examples Pt 3 How many possible postal codes are there in Canada? The format is LDL DLD 26 x 10 x 26 x 10 x 26 x 10 = 17 576 000 263 x 103 How many are in Ontario where the 1st letter is K, L, M, N or P 5 x 10 x 26 x 10 x 26 x 10 = 3 380 000 What if there are no repeated characters? 2160000 5 x 10 x 25 x 9 x 24 x 8

Examples Pt 4 How many 4-digit PINs are possible on a smartphone? 10x10x10x10 = 10 000 What if you saw the first character pressed? 10x10x10 = 1000 If you saw the first character, what is the probability of guessing the PIN correctly? 1 / 1000

Minds on! Suppose Mr. Lieff installs a 4-digit capcha on his website How many different answers are possible? 10x10x10x10 = 10 000 What if no numbers can be repeated? 10x9x8x7 = 5 040 What if you know the first digit is 3? 1x10x10x10 = 1 000 What is the probability of guessing each of these?

Independent and Dependent Events Independent events: the occurrence of one event does not change the probability another Coins, dice, cards w/replacement, spinners are all independent Ex. What is the probability of getting heads when you have thrown an even die? These are independent events, because knowing the outcome of the first does not change the probability of the second

Multiplicative Principle for Probability of Independent Events If we know that A and B are independent events, then… P(B | A) = P(B) Knowing A occurred does not affect P(B) We can also prove that if two events are independent the probability of both occurring is… P(A and B) = P(A) × P(B)

Example 1 A sock drawer has a red, a green and a blue sock You pull out one sock, replace it and pull another out a) Draw a tree diagram representing the possible outcomes b) What is the probability of drawing the red sock both times? These are independent events R B G

Example 2 If you draw a card, replace it and draw another, what is the probability of getting two aces? 4/52 x 4/52 = 16/2704 = 1/169 = 0.006 These are independent events What if the first card is not replaced? 4/52 x 3/51 = 12/2652

Example 3 - Predicting Outcomes What if the outcomes are not equally likely? Mr. Lieff is playing Texas Hold’Em He finds that he wins 70% of the pots when he does not bluff He also finds that he wins 50% of the pots when he does bluff If there is a 60% chance that Mr. Lieff will bluff on his next hand, what are his chances of winning the pot? We will start by creating a tree diagram

Tree Diagram 0.5 Win pot P=0.6 x 0.5 = 0.3 bluff 0.6 0.5 Lose pot 0.7 P=0.4 x 0.7 = 0.28 0.4 no bluff 0.3 Lose pot P=0.4 x 0.3 = 0.12

Continued… P(no bluff, win) = P(no bluff) x P(win | no bluff) P(bluff, win) = P(bluff) x P(win | bluff) = 0.6 x 0.5 = 0.30 Probability of a win: 0.28 + 0.30 = 0.58 So Mr. Lieff has a 58% chance of winning the next pot

MSIP / Home Learning Read the examples on pages 239-244 Complete pp. 245 – 249 #2, 3, 5, 7, 9, 12, 13a, 14