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Math 221

Math 221. Integrated Learning System Week 2 Lecture 2 Conditional Probability, Intersection, and Independence of Events. Conditional Probability. What is the probability of rolling a prime number on a single roll of a die?. Let S = {1, 2, 3, 4 ,5, 6}.

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Math 221

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  1. Math 221 Integrated Learning System Week 2 Lecture 2 Conditional Probability, Intersection, and Independence of Events

  2. Conditional Probability What is the probability of rolling a prime number on a single roll of a die? Let S = {1, 2, 3, 4 ,5, 6} Then the event of rolling a prime number is A = {2, 3, 5}.

  3. Conditional Probability What would be the probability of having rolled a prime number if we know that the number rolled is even? Let the event an even number is rolled be B = {2, 4, 6}.

  4. Conditional Probability

  5. 2 3 4 1 6 5 Example What is the probability of the pointer landing on a number greater than 4?

  6. Example (continued) What is the probability of the pointer landing on a number greater than 4 given that it has landed on an even number?

  7. Example What is the probability of no rain? What is the probability of an accident and no rain? What is the probability of an accident given no rain?

  8. Intersection of Events Solving each equation above for the probability of the intersection: P(AB) = P(B)P(A|B) and P(BA) = P(A)P(B|A) This is called the product rule.

  9. Example If 60% of a department store’s customers are female and 80% of the male customers have a charge account, what is the probability that a customer selected at random is a male and has a charge account? Let A be the event the customer is male. P(A) = 1-.6=.4 Let B be the event that the customer has a charge account. P(B|A) = .8 P(BA) = .4(.8) = .32

  10. Probability Trees Suppose we have 6 blue balls and 4 red balls in a box. If we take 2 balls in succession from the box without replacement, what is the probability of drawing two red balls? (We are assuming a blind draw so that any ball is equally likely to be drawn.)

  11. 1/3 2/15 .4 2/3 4/15 4/15 4/9 .6 1/3 5/9 Probability Trees

  12. Example for You Two balls are drawn in succession without replacement from a box containing 4 red balls and 2 white balls. What is the probability of having a red ball on the second draw?

  13. Constructing a Probability Tree • Draw a tree diagram corresponding to all combined outcomes of the sequence of experiments. • Assign a probability to each tree branch. • Use these results to answer various questions related to the sequence of events as a whole.

  14. Stochastic Processes The process of carrying out a sequence of related experiments where the outcomes of each experiment are uncertain is called a stochastic process.

  15. Example A large computer company A subcontracts the manufacturing of its circuit boards to two companies, 40% to company B, and 60% to company C. Company B in turn subcontracts 70% of the orders it receives from company A to company D and the remaining 30% to company E. When the boards are completed by companies C, D, and E, they are shipped to company A to be used in various computer models. It has been found that 1.5%, 1%, and .5% of the boards from D, E, and C respectively, prove defective during the 90 day warranty period after a computer is first sold. What is the probability that a given board in a completed computer will be defective during the 90 day warranty period?

  16. .985 .2758 .7 .0042 .015 .99 .1188 .4 .3 .0012 .01 .995 .597 .6 .003 .005 Solution to Example

  17. Solution to Example (continued) Probability of a defective board = .0042 + .0012 + .003 = .0084. What is the probability that a circuit board in a completed computer is from company E or C? P(EC) = .4(.3) + .6 = .72

  18. Independent Events Two events A and B are said to be independent if the probability of one event is not affected by the occurrence of the other. Mutually exclusive events are not independent. If you know that one event has occurred, the probability of the other is nil. Definition: If A and B are any events in a sample space S, we say that A and B are independentif and only if P(AB) = P(A)P(B)

  19. Theorem 1 If A and B are independent events with non-zero probability in a sample space S, then P(A|B) = P(A) P(B|A) = P(B)

  20. Example Consider the experiment of flipping a coin twice. The sample space is S = {HH, HT, TH, TT}. Let A be the event a head is the first toss. B be the event a head is the second toss. Compute P(B|A) and compare with P(B). A = {HH, HT}, so P(B|A) = 1/2. B = {HH, TH}, so P(B) = 2/4 = 1/2

  21. Example A single card is drawn from a 52 card deck. Test the following events for independence. (A) E = The drawn card is a red card. F = The drawn card’s number is divisible by 5 (B) G = the drawn card is a king. H = The drawn card is a queen. P(G) = P(H) = 1/13. P(GH) = 0. So the events are dependent.

  22. Independent Set of Events

  23. Example A single die is rolled six times. What is the probability of rolling the sequence 1, 2, 3, 4, 5, 6? Since the die has no memory, the probability of rolling a given number on any roll is not affected by the previous rolls. So, P(1, 2, 3, 4, 5, 6) = P(1)P(2)P(3)P(4)P(5)P(6) = (1/6)6 = 1/46656.

  24. Conditional Probability Product Rule Summary

  25. Independent Events • A and B are independent iff P(AB) = P(A)P(B). • If A and B are independent events with non-zero probability, then P(A|B) = P(A) and P(B|A) = P(B). • If A and B are events with nonzero probability and either P(A|B) = P(A) or P(B|A) = P(B) then A and B are independent • If E1, E2,…En are independent, then P(E1E2...En) = P(E1)P(E2)…P(En)

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