Chapter 6 Probabilit y

Vocabulary Probability – the proportion of times the outcome would occur in a very long series of repetitions (likelihood of an event occuring) Sample space S – set of all possible outcomes Event – an outcome or a set of outcomes Relative frequency – likelihood of occurrence (%) Trials – repetitions

Rolling a Die Sample Space: S = {1, 2, 3, 4, 5, 6} Possible Event: A = rolling a 3P(A) = 1/6 Possible Event: B = rolling an oddP(B) = 3/6 Possible Event: C = number greater than 4 P(C) = 2/6 If we rolled the die 600 times, what type of distribution would we see? Sketch the graph. Probability Model (probability distribution): xi pi 1/6 1/6 1/6 1/6 1/6 1/6

Flipping 2 Coins Sample Space: S = {HH, HT, TH, TT} Possible Event: 2 headsP(2 heads) = 1/4 Possible Event: one head P(1 head) = 2/4 Probability Model (probability distribution): If we flip the coin again, does the result of the first flip influence the result of the 2 nd flip? No, events are independent. p i ¼¼¼¼ x i HHHTTHTT

More Vocabulary Multiplication Principle: if you can do one task in a number of ways and a second task in b number of ways, then both tasks can be done in a ∙ b number of ways. Example: What is the total number of outcomes of flipping 4 coins (either all at once or one coin 4 times)? With Replacement: Draw a card from a deck of cards. Put it back, shuffle and draw again. Without Replacement: Draw a card from a deck of cards. Set it aside and draw another card. Example: Select a random digit by drawing numbered slips of paper from a hat. How many 3 digit numbers can you make a. with replacement? b. without replacement?

More Vocabulary Rolling a fair dice is an example of independent events. Knowledge of what the first roll was has no influence on the next roll. If I were to randomly select students in class of 100 students without replacement, the probability of you being selected changes with each name called. 1/100, 1/99, 1/98,…0. Independent Events: knowing one event has occurred does not change the probability of the other event occurring. Dependent Events: knowing one event has occurred changes the probability of the other event occurring. Conditional Probability

TREE DIAGRAMS flipping a coin and rolling a die H T H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 12 Events (6 x 2) P(Heads and 1) = P(Heads and Even) = P(Tails) = P(Even) =

Examples In a test of a new package design, you drop a package of 4 glass ornaments. 1.Describe the sample space. 2.Given the probability distribution table of your results, calculate the probabilities: a.Complete the table. b.What is the probability of breaking at least 2 glasses? c. What is the probability of breaking no more than 3 glasses? xixi pipi ?

Summary of Rules 0  P(event)  1 P(S) = 1, where S is Sample Space P(event C ) = 1 – P(event) P(A or B) = P(A) + P(B) (if A and B disjoint) P(A or B) = P(A) + P(B) – P(A and B) P(A and B) = P(A)P(B) (if A and B independent)

Conditions for Valid Probabilities 0  P(event)  1 The probability of an event must be between 0 and 1. P(tails) =.5 P(roll a 4) = 1/6P(today is Friday) = _?_ P(S) = 1 Sum of all probabilities of events in a sample space is 1. P(heads) + P(tails) = 1P(weekday) + P(weekend) = 1 P(A) + P(A C ) = 1 A C is the complement of event A. If A = heads, then A C = tails If B = heart, then B C = not heart

More Vocabulary Two events are DISJOINT or MUTUALLY EXCLUSIVE if they do not contain any of the same events and can not occur simultaneously. Joint Events: the simultaneous occurrence of two events Joint Probability: the probability of a joint event occurring Independent Events: knowing one event has occurred does not change the probability of the other event occurring. Dependent Events: knowing one event has occurred changes the probability of the other event occurring. Disjoint events cannot be independent!

Union of 2 Events: P(A or B) = P(A B) If two events are disjoint (or mutually exclusive) P(A B) = P(A or B) = P(A) + P(B) For ALL events A and B P(A B) = P(A or B) = P(A) + P(B) – P(A and B) ∩ ∩ ∩ Intersection of 2 Events: P(A and B) = P(A∩B) If independent: P(A∩B) = P(A and B) = P(A) P(B) If dependent: Use common sense probability or conditional probability. If disjoint: P(A and B) = {} or Ø (this is the empty set)

1. Find the probability that you draw either an ace or a red card. P(Ace or Red) = P(Ace) + P(Red) – P(Red Ace) = (4/52) + (26/52) – (2/52) = 28/52 2. Find the probability of rolling 2 sixes. P( roll 6 and roll 6) = P(roll 6)P(roll 6) = (1/6)(1/6) = 1/36

3. Find the probability of drawing a 7  and a 6 w/o replacement. P( 7  and 6 ) = = P(7  )P(6 given 7  was drawn) = (1/52)(1/51) 4.Find the probability of rolling a 7 or a 10. P(7 or 10) = P(7) + P(10) – P(7 and 10) = (6/36) + (3/36) = (9/36)

For ANY 2 events: P(B|A) = P(A∩B) P(A) P(A∩B) = P(A) P(B|A) Two events are independent if: P(B|A) = P(B) (cannot use multiplication rule)

Probability of one event GIVEN another has occurred P(B|A) = P(A∩B)(“probability of B given A”) P(A) P(A∩B) = P(A) P(B|A) Or you can use common sense!

Conditional Probability Examples Common Sense! 1.Find the probability of drawing a red ball out of a box containing 3 red, 4 blue, and 1 white GIVEN that a blue ball has been drawn and not replaced. P(R|B) = 3/7

Conditional Probability Tree Diagrams! 2. During a one-and-one shooting foul in a basketball game, what is the probability that an 80% free throw shooter makes both baskets? make (.8) miss (.2) 1 st shot make (.8) miss (.2) 2 nd shot 2pts =.8*.8=.64 1pt =.8*.2=.16 0pts =.2 Answer: P(making both shots) =.64

3. The probability of having a certain disease is.05. The probability of testing + if you have the disease is.98; the probability of testing + when you do not have it is.10. What is the probability that you have the disease if you test +? Conditional Probability Examples The Formula!

Are the following events disjoint? Are the following events independent? 1.Drawing a jack and a king. 2.Drawing a red card and a king. 3.Drawing an even card and a face card. 4.Drawing an ace and a black card. 5.Drawing an ace and a queen. 6.Drawing a diamond and a red card.