Copyright © 2016, 2013, and 2010, Pearson Education, Inc. 9 Chapter Probability Copyright © 2016, 2013, and 2010, Pearson Education, Inc.

9-2 Multistage Experiments and Modeling Games Students will be able to understand and explain: Multistage experiments; Independent events; Conditional probability; and Modeling games.

Tree Diagrams Suppose the spinner is spun twice. The tree diagram shows the possible outcomes.

Tree Diagrams The sample space can be written {BB, BR, BY, RB, RR, RY, YB, YR, YY}. The probability of each outcome is

Tree Diagrams Alternatively, we can generate the sample space using a table. YB

Example Suppose we toss a fair coin 3 times and record the results. Find each of the following: a. The sample space for this experiment. S = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT} b. The event A of tossing 2 heads and 1 tail A = {HHT, HTH, THH}

Example (continued) c. The event B of tossing no tails. B = {HHH} d. The event C of tossing a head on the last toss C = {HHH, HTH, THH, TTH}

Example a. When rolling a pair of fair dice, find the probability of the event A = rolling double sixes. There are 36 possible rolls. There is only one way to roll double sixes.

Example (continued) b. Find the probability of rolling a sum of 7 or 11 when rolling a pair of fair dice. There are 36 possible rolls.

Example (continued) There are 6 ways to form a sum of “7”, so

Example (continued) The sample space for the experiment is {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, but the sample space is not uniform; i.e., the probabilities of the given sums are not equal. The probability of rolling a sum of 7 or 11 is

The Sum of 2 Dice: Bar Graph

Example A fair pair of dice is rolled. Let E be the event of rolling a sum that is an even number and F the event of rolling a sum that is a prime number. Find the probability of rolling a sum that is even or prime, that is, P(E U F). E U F = {2, 4, 6, 8, 10, 12, 3, 5, 7, 11}

Example (continued)

Example (continued) Alternate solution 1: E = {2, 4, 6, 8, 10, 12} and F = {2, 3, 5, 7, 11}. Thus, E and F are not mutually exclusive because E ∩ F = {2}.

Example (continued) Alternate solution 2: E = {2, 4, 6, 8, 10, 12} and F = {2, 3, 5, 7, 11}. Thus, E U F = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12} and E U F = {9}.

More Multistage Experiments The box in figure (a) contains one black ball and two white balls. If a ball is drawn at random and the color recorded, a tree diagram for the experiment might look like the one in figure (b). Because each ball has the same chance of being drawn, we may combine the branches and obtain the tree diagram shown in figure (c).

Suppose a ball is drawn at random from the box in the figure and its color recorded. The ball is then replaced, and a second ball is drawn and its color recorded. The sample space for this two-stage experiment may be recorded using ordered pairs as

The tree diagram for this experiment is shown below.

Multiplication Rule for Probabilities for Tree Diagrams For all multistage experiments, the probability of the outcome along any path of a tree diagram is equal to the product of all the probabilities along the path.

Suppose two balls are drawn one-by-one without replacement.

Independent Events Independent events: When the outcome of one event has no influence on the outcome of a second event.

Independent Events If two coins are flipped and event E1 is obtaining a head on the first coin and E2 is obtaining a tail on the second coin, then E1 and E2 are independent events because one event has no influence on possible outcomes.

Independent Events

Independent Events For any independent events E1 and E2,

Example The figure shows a box with 11 letters. Some letters are repeated. Suppose 4 letters are drawn at random from the box one-by-one without replacement. What is the probability of the outcome BABY, with the letters chosen in exactly the order given?

Example (continued) We are interested in only the tree branch leading to the outcome BABY. The probability of the first B is because there are 2 B’s out of 11 letters. The probability of the second B is because there are 9 letters left after 1 B and 1 A have been chosen.

Example Consider the three boxes shown below. A letter is drawn from box 1 and placed in box 2. Then, a letter is drawn from box 2 and placed in box 3. Finally, a letter is drawn from box 3. What is the probability that the letter drawn from box 3 is B? (Call this event E.)

Example (continued)

Conditional Probabilities If A and B are events in sample space S and P(A)  0, then the conditional probability that an event B occurs given that event A has occurred is given by Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

Example What is the probability of rolling a 6 on a fair die if you know that the roll is an even number? If event B is rolling a 6 and event A is rolling an even number, then Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

Modeling Games There are two black marbles and one white marble in a box. Gwen mixes the marbles, and Arthur draws two marbles at random without replacement. If the two marbles match, Arthur wins; otherwise, Gwen wins. Does each player have an equal chance of winning?

Modeling Games The probability that the marbles are the same color is while the probability that they are not the same color is Thus, the players do not have the same chance of winning.