4/17/2012. NE ( ) # of ways an event can happen NS ( ) # of events in the sample space

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I. Vocabulary: A. Outcomes: the things that can happen in a probability experiment B.

Probability of an Event (P(E)): the likelihood that an event occurs NE ( ) # of ways an

Relative Frequency: the probability given by an experiment # of successes Relative Frequency

If E is empty, then P(E) = 0. III. Examples A. Example : A fair coin is tossed three times.

T HHH HHT HTH HTT THH THT TTH TTT N( head) 3 P( head) = = N(sample space) 8 B.

Which number has the 3 4 5 7 greatest probability?

1 I. Vocabulary: A. Outcomes: the things that can happen in a probability experiment B. Sample Space (S): all possible outcomes C. Event (E): one outcome D. Probability of an Event (P(E)): the likelihood that an event occurs NE ( ) # of ways an event can happen PE ( ) = = NS ( ) # of events in the sample space E. Relative Frequency: the probability given by an experiment # of successes Relative Frequency = # of trials II. Basic Properties of Probability A. 0 PE ( ) B. If E = S, then P(E) = C. If E is empty, then P(E) = 0. III. Examples A. Example : A fair coin is tossed three times. What is the probability of getting exactly one head H Sample Space: H T H H T T H H T T T H T HHH HHT HTH HTT THH THT TTH TTT N( head) 3 P( head) = = N(sample space) 8 B. Find the sample space for finding the sum on two six-sided dice Which number has the greatest probability? , (7) 3 = What is the probability of rolling a number greater than 8? P(9,0,,) = = = Class work: Lesson Master Worksheet Homework: p # 4-7, 9, 3

I. Set Notation: A. A B The Intersection of Sets A and B (the objects in both sets A and B) B. A B the Union of sets A and B (the objects in either A or B) II. Addition Counting Principle: A.

If we divide id by the sample space, we get the probability equivalent. N ( A B ) ( ) ( ) = N A + N B NS ( ) NS ( ) NS ( ) 3.

2 I. Set Notation: A. A B The Intersection of Sets A and B (the objects in both sets A and B) B. A B the Union of sets A and B (the objects in either A or B) II. Addition Counting Principle: A. Mutually Exclusive Form (For mutually exclusive sets, there is no overlap). For two mutually exclusive sets A and B: N( A B) = N( A) + N( B). If we divide id by the sample space, we get the probability equivalent. N ( A B ) ( ) ( ) = N A + N B NS ( ) NS ( ) NS ( ) 3. Example : What is the probability of tossing an even number or 3 when rolling a die? Why are these mutually exclusive? there is no overlap P(even or 3) = P( even) + P(3) P (even or 3) = = + = = 3 P( A B) = P( A) + P( B) B. Addition Counting Principle General Form N( A B) = N( A) + N( B) N( A B) C. Probability Version PA ( B ) = PA ( ) + PB ( ) PA ( B ) D. Example : 3 states contain land west of the continental divide, but 4 states contain land east of the divide. How many states have land on both sides? N( E W) = N( E) + N( W) N( E W) 50 = N ( E W ) 5 = N( E W) 5 = N( E W)

3 III. Complementary Events A. The complement of set A is called not A or A. P(not A) = P( A) = P( A) Class work: 7. lesson master worksheet Homework: p #, 3, 5,, 8 4. B. Example 3: Find the complement for each of the following events.. Flipping i two tails on two coins flipping less than tails HH, HT, TH. Rolling a 5 on one die rolling any number except 5,,,3,4, 3. Drawing a spade in a deck of cards. drawing hearts, diamonds or clubs

I. With Replacement A. With Replacement you can get the same answer multiple times.. Example : Suppose you have a question true and false quiz.

Theorem: For a set with n elements, there are n k possible arrangements of k elements with replacement. 3.

Without Replacement A. Without Replacement You cannot get the same answer more than once B.

Example 4: Evaluate each of the following. 8! = 40,30. 4!.3 0 89 C. Theorem: There are n! arrangements of n elements without replacement 3. 87!/85!

4 I. With Replacement A. With Replacement you can get the same answer multiple times.. Example : Suppose you have a question true and false quiz. How many arrangements of answers are possible? = = 4. Theorem: For a set with n elements, there are n k possible arrangements of k elements with replacement. 3. Example : what is the probability of getting a perfect score by guessing? N(perfect papers) P(perfect score)= = N (possible papers) 4 II. Without Replacement A. Without Replacement You cannot get the same answer more than once B. Example 3: How many ways can you answer a matching test with questions (if each answer can only be used once) 543 =! = 70 III. Examples A. Example 4: Evaluate each of the following. 8! = 40,30. 4! C. Theorem: There are n! arrangements of n elements without replacement 3. 87!/85! = = 87 8 B. Example 5: Suppose the license plates in a state have six letters or numbers (excluding I and O). How many different license plates are possible assuming letters can be repeated? CW: LM 7-3 Homework: p #3, 5-4, -8, 0- There are 34 different characters that can be selected (0 numbers and 4 letters) = 34 =,544,804,4

7.4: Permutations I. Permutations: A. Each way that a set of objects can be arranged is called a permutation. For example, ABC, ACB, and BCA are all permutations of the letters A, B, and C. B. There are n!

Example : If there are ten teams in a given conference, how many different possible standings can result (assuming no ties). 0 9... 3 = 0! = 3,8,800 II. Permutations of n objects taken r at a time. A.

5 7.4: Permutations I. Permutations: A. Each way that a set of objects can be arranged is called a permutation. For example, ABC, ACB, and BCA are all permutations of the letters A, B, and C. B. There are n! permutations of n different elements. There are 3!= permutations of the letters A, B, and C. ABC, ACB, BAC, BCA, CAB, CBA C. Example : If there are ten teams in a given conference, how many different possible standings can result (assuming no ties) = 0! = 3,8,800 II. Permutations of n objects taken r at a time. A. Example : in the same conference from example, how many ways can 4 teams be selected to make it to the playoffs? B. We are making groups of 4 out of groups of = 5,040 first place second third fourth C. Formula for n objects taken r at a time( npr ). The number of permutations of n taken r at a time is: n r ( )( )... ( ) P = n n n n r+ = n!! ( n r). Example 3: Use the formulas above to calculate the answer from example : 0( 0 )( 0 )( 0 3) = = 5,040 0! 0! = = !! ( ) D. Example : how many different four letter words can be selected from the word equations (the words do not need to make sense, and cannot have repeated letters). We are choosing 4 of 9 9P4 = 987 = 3,04words

6 E. Note: if all n objects are selected, then the alternate formula yields: n! n! n! np = n n! n n! = 0! = = ( ) for this to work: 0! = III. Examples A. In a 4 person relay race, the fastest runner is typically the last person to run. If that place is set, how many different orders are there for the other runners? 3 P3 = 3! = orderings B. Solve the following: P = 4 n 4 ( np ) ( )( )( 3) = 4 ( ) ( n )( n 3) = 4 n n n n n n n 5n+ = 4 n 5 n 3 = 0 ( n 9)( n+ 4) = 0 n = 9, 4 CW: LM 7-4 HW: p #3,, -5, 8,, 4 check: ( P ) P = 4 =

7 I. Combinations A. Permutations: order matters (AB BA) 8.: Combinations. Three students are chosen to put problems through 3 on the board. Nine players are chosen for a person lineup on a baseball team B. Combinations: order does not matter (AB=BA). Three toppings are chosen from eight for a pizza. A 4 person committee is chosen from a class. C. Example : how many ways can 5 people shake hands with everyone exactly once? People A, B, C, D, and E A B C D E 54 = 0 Note the repeats A BA CA DA EA B AB CB DB EB C AC BC DC EC D AD BD CD ED E AE BE CE DE D. Formula: n n! = nc = r r ( n r)! r! Note that we divide the permutation formula by r! in order to divide out the repeats 0 0 = II. Examples: A. Example : Six points are chosen on a circle. How many line segments can be drawn using these points as endpoints? C = 5. How many triangles are there with these points as vertices? C 3 = 0 3. How many quadrilaterals are there with these points as vertices? C 4 = 5 B. Example 3: you order a pizza with 9 toppings. How many different pizzas are possible with exactly 3 toppings? 9C 3 = 84

8 C. Five students from your class of 5 are randomly chosen to be interviewed. What is the probability that neither you nor your best friend in the class will be chosen? Number of ways to choose 5 students without choosing you or your friend: 3 C 5 = 33,49 Number of ways to choose 5 from the whole class: 5C 5 = 53,30 33,49 Probability = ,30 CW: LM 8. HW: p # -9, -9, 0 (a,b,c), 4.

I. Independent Events A. Two events are Independent iff one event does not make the other event more or less probable. B.

Selections with replacement 3. Rolling a 5 on one die and the chance that it rains D.

Find the probability that if you draw two marbles (with replacement) and they are both red.

Now, find the sample space and then find P ( A B ) R R B A B = { RR, RR, RR, RR} R RR RR BR R RR RR BR N( A B) 4 PA ( B) =

4 4 PA ( B) =, PA ( ) PB ( ) = = 9 3 3 9 P( A B) = P( A) P( B) yes II. Dependent Events A.

9 I. Independent Events A. Two events are Independent iff one event does not make the other event more or less probable. B. This is true when: P( A B) = P( A) P( B) C. Examples of Independent Events:. Flipping a fair coin twice. Selections with replacement 3. Rolling a 5 on one die and the chance that it rains D. Example : A bag contains three marbles: red and blue (we will label them R, R, and B). Find the probability that if you draw two marbles (with replacement) and they are both red.. Let A be the event that the first marble is red and B be the event that the second marble is red. PA= ( ) PB ( ) = 3 3. Now, find the sample space and then find P ( A B ) R R B A B = { RR, RR, RR, RR} R RR RR BR R RR RR BR N( A B) 4 PA ( B) = = NS ( ) 9 B RB RB BB 3. Are these two events independent? 4 4 PA ( B) =, PA ( ) PB ( ) = = P( A B) = P( A) P( B) yes II. Dependent Events A. Two events are dependent iff one event changes the probability of the other event. B. This occurs when: P( A B) P( A) P( B) C. We will repeat Example, without replacement R R B A B = { RR R RR RR BR, RR} R RR RR BR N( A B) PA ( B) = = = NS ( ) 3 B RB R B BB D. Are these events independent? 4 PA ( B) =, PA ( ) PB ( ) = = P( A B) P( A) P( B) no

10 III. Examples: A. A fair coin is tossed 4 times. Let A= all heads and B=all tails. Are A and B independent or dependent? PA= ( ) = 4 PB ( ) = = 4 No : P( A) P( B) P( A B) PA ( B) = 0 B. Two normal dice are tossed. Let C=the sum is 7 and D=the first die is 5. Are C and D independent or dependent? C = { +,+,+ 5,5+,3+ 4,4+ 3} D = { 5 +,5 +,5 + 3,5 + 4,5 + 5,5 + 5} C D= { 5+ } PC ( ) = = 3 PD ( ) = 3 = PA ( B) = 3 Yes : P( A) P( B) = P( A B) CW: Lesson Master 7-5 HW: HW: p #, 9, 0, -4, -

11 I. Probability Distributions. A random variable is a variable that takes on the values from a probability experiment.. A Probability Distribution is a function that maps values from a random variable onto its probability. I. Example : Consider the probabilities when you toss two die and look at the sum of the numbers: x P(x)= /3 /3 3/3 4/3 5.3 /3 5/3 4/3 3/3 4/3 5/3 x is a random variable P(x) This is a graph of the probability distribution x II. Expected Value (Mean) A. In order to calculate the expected value or mean, you multiply each value for the random variable by its probability and add them up. n μ = ( xipx ( i)) i= B. Example : Calculate the expected value of the random variable from Example : 3 μ = μ = 7 Classwork: probability distribution assignment. Homework: p # 9,,, 4,

LECTURE NOTES by DR. J.S.V.R. KRISHNA PRASAD

LECTURE NOTES by DR. J.S.V.R. KRISHNA PRASAD .0 Introduction: The theory of probability has its origin in the games of chance related to gambling such as tossing of a coin, throwing of a die, drawing cards from a pack of cards etc. Jerame Cardon,