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Chapter 6 Probability

Chapter 6 Probability. Vocabulary. Probability – the proportion of times the outcome would occur in a very long series of repetitions (l ikelihood of an event occuring ) Sample space S – set of all possible outcomes Event – an outcome or a set of outcomes

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Chapter 6 Probability

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  1. Chapter 6 Probability

  2. 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

  3. Rolling a Die Sample Space: S = {1, 2, 3, 4, 5, 6} Possible Event: A = rolling a 3 P(A) = 1/6 Possible Event: B = rolling an odd P(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): xi1 2 3 4 5 6 pi1/6 1/6 1/6 1/6 1/6 1/6

  4. Flipping 2 Coins Sample Space: S = {HH, HT, TH, TT} Possible Event: 2 heads P(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 2nd flip? No, events are independent. xi HH HT TH TT pi ¼ ¼ ¼ ¼

  5. 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)? More Vocabulary 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?

  6. 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 More Vocabulary

  7. TREE DIAGRAMSflipping a coin and rolling a die H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 1 2 3 4 5 6 1 2 3 4 5 6 12 Events (6 x 2) P(Heads and 1) = P(Heads and Even) = P(Tails) = P(Even) = H T

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

  9. Summary of Rules • 0  P(event)  1 • P(S) = 1, where S is Sample Space • P(eventC) = 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)

  10. 0  P(event)  1 The probability of an event must be between 0 and 1. P(tails) = .5 P(roll a 4) = 1/6 P(today is Friday) = _?_ P(S) = 1 Sum of all probabilities of events in a sample space is 1. P(heads) + P(tails) = 1 P(weekday) + P(weekend) = 1 P(A) + P(AC) = 1 AC is the complement of event A. If A = heads, then AC = tails If B = heart, then BC = not heart Conditions for Valid Probabilities

  11. 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! More Vocabulary

  12. ∩ 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)

  13. Examples 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

  14. Examples 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) • 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)

  15. Summary of Rules Conditional Probability 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)

  16. Conditional Probability 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! same formula! Remember – you can only use P(A∩B) = P(A)P(B) if independent!

  17. Conditional Probability ExamplesCommon Sense! • 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

  18. Conditional ProbabilityTree 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? 2pts = .8*.8=.64 make (.8) 2nd shot make (.8) miss (.2) 1st shot 1pt = .8*.2=.16 miss (.2) 0pts = .2 Answer: P(making both shots) = .64

  19. Conditional Probability ExamplesThe Formula! 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 +?

  20. Are the following events disjoint?Are the following events independent? • Drawing a jack and a king. • Drawing a red card and a king. • Drawing an even card and a face card. • Drawing an ace and a black card. • Drawing an ace and a queen. • Drawing a diamond and a red card.

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