# Aim: Probability! Probability! - How do I count the ways?

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Aim: Probability! Probability! - How do I count the ways?
Do Now: Choose one item from each category to make an entire meal. Main Course Drink Dessert Spaghetti Milk Ice Cream Hamburger Coke Apple Pie Hotdog Chocolate Cake

Tree Diagram Dinner is Served! Spaghetti Hamburger Hotdog
Main Course Drink Dessert Sample Space Ice Cream S M I Milk Apple Pie S M A Chocolate Cake S M C Spaghetti Hamburger Hotdog Ice Cream Apple Pie Chocolate Cake S C I S C A S C C Coke Ice Cream Apple Pie Chocolate Cake Milk Coke H M A Ht M I Ht M A Ht M C H M I H M C H C I H C A H C C Ht C I Ht C A Ht C C Tree Diagram Three consecutive Events

Am I lost? How many different ways will get us from MJ Petrides to Great Adventure? MJ Petrides 3 Outerbridge Crossing Tracing the different routes we find there are 6 different routes. 2 Is there a shortcut method for finding how many different routes there are? Great Adventure

The Fundamental Counting Principle
To find the total number of possible outcomes in a sample space, multiply the number of choices for each stage or event... in other words... If event M can occur in m ways and is followed by event N that can occur in n ways, then the event M followed by event N can occur in m · n ways. Counting Principle events: m · n events: m · n · o events: m · n · o · p events: etc.

18 x = 3 2 3 Spaghetti Hamburger Hotdog Main Course Drink Dessert
Sample Space Ice Cream S M I Milk Apple Pie S M A Chocolate Cake S M C Spaghetti Hamburger Hotdog Ice Cream Apple Pie Chocolate Cake S C I S C A S C C Coke Ice Cream Apple Pie Chocolate Cake Milk Coke H M A Ht M I Ht M A Ht M C H M I H M C H C I H C A H C C Ht C I Ht C A Ht C C

12 outcomes in sample space
Model Problem Jamie has 3 skirts - 1 blue, 1 yellow, and 1 red. She has 4 blouses - 1 yellow, 1 white, 1 tan and 1 striped. How many skirt-blouse outfits can she choose? What is the probability she will chose the blue skirt and white blouse? 3 4 Skirt blouse 12 outcomes in sample space yellow white tan striped b y b w b t b s Blue yellow white tan striped y y y w y t b y Yellow yellow white tan striped r y r w r t r y Red

Regents Question A four-digit serial number is to be created from the digits 0 through 9. How many of these serial numbers can be created if 0 can not be the first digit, no digit may be repeated, and the last digit must be 5? 1) ) ) ) 2,520 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 10 possible outcomes to start possible outcomes 8    8 7 1 = 448 E1 E2 E3 E4

Types of Events Compound Event - two or more activities Ex. Rolling a pair of dice What is the probability of rolling a pair of dice and getting a total of four? Die 1 P(4) = 3/36 = 1/12 Die 2

Types of Events Mutually Exclusive Events – Two or more events that can not happen at the same time Ex. mutually exclusive? rolling a 2 and a 3 on a die rolling an even number or a multiple of 3 on a die Yes No Independent Event – when the outcome of one event does not affect the outcome of a second event. Dependent Event - when the outcome of one event affects the outcome of a second event.

Probability of Two Independent Events
The probability of two independent events can be found by multiplying the probability of the first event by the probability of the second event (The Counting Principle w/Probabilities) AND Probabilities P(A and B) = P(A) · P(B) Ex: A die is tossed and a spinner is spun. What’s the probability of throwing a 5 and spinning red? P(5 and R)? Faster than drawing a tree diagram!! P(5 and Red) = Independent Event

ONLY IF THE FIRST PINK CHIP WAS NOT RETURNED TO THE BAG. Event A
Not So Independent! There are 4 red, 3 pink, 2 green and 1 blue chips in a bag. What is P(pink)? 3/10 What is the probability of picking a pink and then reaching in and picking a second pink without replacing the first picked pink? ONLY IF THE FIRST PINK CHIP WAS NOT RETURNED TO THE BAG. Event A Event B P(pink) P(pink) 3/10 2/9 P(pink and pink) = The selection of the second event was affected by the selection of the first. Dependent Event

Model Problem Find the probability of choosing two pink chips without replacement. P(pink) Event A Event B P(pink) Dependent Events P(pink, pink) = 3/10 3/10 • 2/9 = 6/90 or 1/15 Counting Principle w/Probabilities 2/9

Model Problem Find the probability of choosing blue and then a red chip without replacement. Dependent Events P(blue, red) = P(blue) Event A Event B P(red) 1/10 4/9 1/10 • 4/9 = 4/90 or 2/45 Counting Principle w/Probabilities

extends to multiple dependent events
Two events are dependent events if the occurrence of one of them has an effect on the probability of the other. AND Probabilities with Dependent Events If A and B are dependent events, then P(A and B) = P(A) · P(B given that A has occurred) extends to multiple dependent events You are dealt three cards from a 52-card deck. Find the probability of getting 3 hearts. P(1st heart) = 13/52 P(2nd heart) = 12/51 P(3rd heart) = 11/50 P(hearts) = 13/52 · 12/51 · 11/50 = 1716/ 

Probability of Dependent Events
1. Calculate the probability of the first event Calculate the probability of the second event, etc. ... but NOTE: The sample space for the probability of the subsequent event is reduced because of the previous events. 3. Multiply the the probabilities. Ex. A bag contains 3 marbles, 2 black and one white. Select one marble and then, without replacing it in the bag, select a second marble. What is the probability of selecting first a black and then a white marble? Key words - without replacement

Model Problem From a deck of 10 cards (5 ten-point cards, 3 twenty-point cards, and 2 fifty-point cards), Ronnie can only pick 2 cards. In order to win the game, he must pick the 2 fifty-point cards. What is the probability that he will win? 10 20 50

Counting Principle w/Probabilities
Model Problem From a deck of 10 cards Ronnie can only pick 2 cards. In order to win the game, he must pick the 2 fifty-point cards. What is the probability that he will win? 10 10 10 10 10 20 20 20 50 50 50 Dependent P(50, 50) = Event A Event B P(50) = 2/10 P(50) = 1/9 2/10 • 1/9 = 2/90 = 1/45 Counting Principle w/Probabilities

A. What is the probability that Penny pulled 3 red balls?
Regents Question Penny has 3 boxes, each containing 10 colored balls. The first box contains 1 red ball and 9 white balls, the second box contains 3 red balls and 7 white balls, and the third box contains 7 red balls and 3 white balls. Penny pulls 1 ball out of each box. Box 1 Box 2 Box 3 A. What is the probability that Penny pulled 3 red balls? P(r,r,r) = 1/10 • 3/10 • 7/10 = 21/1000 B. If Penny pulled 3 white balls and did not replace them, what is the probability that she will now pull 3 red balls? P(r,r,r) = 1/9 • 3/9 • 7/9 = 21/729

Venn Diagrams Find the probability of rolling a die and getting a number that is both odd and greater than 2. odd > 2 1 3 5 4 6 3 5 In logic, a sentence p and q, written p  q, is true only when p is true and q is true. P(odd) = 3/6 P(> 2) = 4/6 2 1 3 5 4 6 P(odd  > 2) = 2/6

Probability of A and B Probability of (A  B)
(A and B are separate events) Example: Find the probability of rolling a die and getting a number that is both odd and greater than 2. In probability, an outcome is in event (A and B) only when the outcome is in event A and the outcome is also in event B.

Mutually Exclusive Events
Mutually exclusive – two events A & B are mutually exclusive if they can not occur at the same time. That is, A and B are mutually exclusive when A  B =  An outcome for A or B is in one or the other. If the events are mutually exclusive then P(A or B) = P(A) + P(B) If one card is randomly selected from a deck of cards, what is the probability of selecting a king or a queen? mutually exclusive? yes

‘Or’ Probabilities Not Mutually Exclusive
From a standard deck you randomly select one card. What is the probability of selecting a diamond or a face card? mutually exclusive? no P(or fcd) = common elements A  B n(A  B) = 3 {K, Q, J}

Mutually Exclusive or Not
1. A card is drawn from a standard deck of 52. Find P(king or queen) mutually exclusive P(king) = P(queen) = P(king or queen) = 2. A card is drawn from a standard deck of 52. Find P(king or face card) not mutually exclusive P(king) = P(face) = P(king or face) = In #1 the 2 events have no common elements. They are mutually exclusive. In #2 a card can be both face and king. They are not mutually exclusive.

Count the number of successes for n > 8 n - odd 9, 10, 11, 12 4
Probability of A or B What is the probability of spinning a number greater than 8 or an odd number? Count the number of successes for n > 8 n - odd 9, 10, 11, 12 4 1, 3, 5, 7, 9, 11 6 not mutually exclusive

Probability of (A or B) P(A or B) = P(A) + P(B) - P(A and B) P(A  B) = P(A) + P(B) - P(A  B) P(A  B) = n(A) + n(B) - n(A  B) n(S) n(S) n(S) If A and B are not mutually exclusive events, then Example: Find the probability of rolling a die and getting a number that is odd or greater than 2. successes {1,3,5} {3,4,5,6}

Probability Rules 1. The probability of an impossible event is 0. 2. The probability of an event that is certain to occur is 1. 3. The probability of an event E must be greater than or equal to 0 and less that or equal to 1. 4. P(A and B) = n(A  B) n(S) 5. P(A or B) = P(A) + P(B) - P(A  B) 6. P(Not A) = 1 - P(A) 7. The probability of any even is equal to the sum of the probabilities of the singleton outcomes in the event. 8. The sum of the probabilities of all possible singleton outcomes for any sample space must always equal 1.

A red king must be red and a king 2 P(red and king) = 52
Model Problems In drawing a card from the deck at random, find the probability that the card is: A. A red king B. A 10 or an ace C. A jack or a club P(A and B) = P(A) · P(B) A red king must be red and a king 2 P(red and king) = 52 mutually exclusive P(A  B) = P(A) + P(B) - P(A  B) 10’s and aces have no common outcomes P(10’s or aces) = 4 52 + _ = 8 not mutually exclusive P(A  B) = P(A) + P(B) - P(A  B) There are 4 jacks and 13 clubs, but one of the cards is both (jack of clubs) P(jacks or clubs) = 4 52 13 + 1 _ 16 =

P(A  B) = P(A) + P(B) - P(A  B)
Model Problems Based on the table below, if one person is randomly selected from the US military, find the probability that this person is in the Army or is a woman. P(A  B) = P(A) + P(B) - P(A  B) not mutually exclusive Active Duty US Military Personnel, in 000’s Air Force Army Marines Navy Total Male 290 400 160 320 1170 Female 70 10 50 200 360 470 170 370 1370 P(Army  Female) = P(A) + P(F) - P(A  F)

Model Problems Five more men than women are riding a bus as passengers. The probability that a man will be the first passenger to leave the bus is 2/3. How many passengers on the bus are men, and how many are women? x = number of women 5 x + 5 = number of men 10 2x + 5 = number of passengers Number of men P(man) = Number of passengers 2 3 x + 5 2x + 5 = 4x = 3x + 15 x = 5 x + 5 = 10

Model Problem A special family has had nine girls in a row. Find the probability of this occurrence. Having a girl is an independent event with P(1 girl) = 1/2 P(A and B) = P(A) · P(B) Probability of two Independent Events extends to multiple independent events

P(A) + P(~A) = 1; P(A) = 1 – P(~A); P(~A) = 1 – P(A)
Model Problem If the probability that South Florida will be hit by a hurricane in any single year is 5/19 What is the probability that S. Florida will be hit by a hurricane in three consecutive years? What is the probability that S. Florida will not be hit by a hurricane in the next ten years? The probability of event (A) plus the probability of "not A” or ~A, equals 1: P(A) + P(~A) = 1; P(A) = 1 – P(~A); P(~A) = 1 – P(A)

Model Problem Three people are randomly selected, one person at a time, from 5 freshman, two sophomores, and four juniors. Find the probability that the first two people selected are freshmen and the third is a junior. P(1st selection is freshman) = 5/11 P(2nd selection is freshman) = 4/10 P(3rd selection is junior) = 4/9 P(F, F, J) = 5/11 · 4/10 · 4/9 = 8/99

Model Problems A sack contains red marbles and green marbles. If one marble is drawn at random, the probability that it is red is 3/4. Five red marbles are removed from the sack. Now, if one marble is drawn, the probability that it is red is 2/3. How many red and how many green marbles were in the sack at the start? x = original red marbles 5 15 y = original number of green marbles 3 4 x__ x + y 2 3 x - 5 x + y - 5 = = 3x + 3y = 4x 2x + 2y - 10 = 3x - 15 3y = x 2y + 5 = x 3y = 2y + 5 y = 5 3y = x = 15

Model Problem Several players start playing a game with a full deck of 52 cards. Each player draws two cards at random, one at a time, from the full deck. Find the probability that a player does not draw a pair. 1st card 2nd card

1 3 5 4 6 3 5 2 1 3 5 4 6 Model Problem odd > 2 P(odd) = 3/6
Find the probability of rolling a die and getting a number that is odd or greater than 2. odd > 2 1 3 5 4 6 3 5 P(odd) = 3/6 P(> 2) = 4/6 2 A  B = {1, 3, 4, 5, 6} 1 3 5 4 6 n(A  B) = 5 n(U) = 6

Model Problem In a group of 50 students, 23 take math, 11 take psychology, and 7 take both. If one student is selected at random, find the probability that the student takes math or psychology P(A  B) = P(A) + P(B) - P(A  B) 23 4 16 7 M Psy

Determine the number of outcomes:
4 coins are tossed A die is rolled and a coin is tossed A tennis club has 15 members: 8 women and seven men. How many different teams may be formed consisting of one woman and one man on each team? A state issues license plates consisting of letters and numbers. There are 26 letters, and the letters may be repeated on a plate; there are 10 digits, and the digits may be repeated. The how many possible license plates the state may issue when a license consists of: 2 letters, followed by 3 numbers, 2 numbers followed by 3 letters.

Model Problem One bag contains 3 red and 4 white balls. A 2nd bag contains 6 yellow and 3 green balls. One ball is drawn from each bag. Find the probability of choosing a red and yellow ball.

The Product Rule

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