Do you know when Exam 1 is?

1. Do you know when Exam 1 is? • Yes • No • I need a break.

2. Upcoming In Class • Homework 5 due Sunday 2/17 • Quiz 3 next Thursday – February 21st • (HW4 and HW5) • Exam 1 – March 7th

3. Chapter 15 Probability Rules!

4. The General Addition Rule • For disjoint events : P(AorB) = P(A) + P(B) • For any two events A and B, P(AorB) = P(A) + P(B) – P(AandB)

5. The General Multiplication Rule • For independent events P(AandB) = P(A) x P(B) • For any two events A and B, P(AandB) = P(A) x P(B|A)

6. Independent ≠ Disjoint • Disjoint events cannot be independent! Well, why not? • Since we know that disjoint events have no outcomes in common, knowing that one occurred means the other didn’t. • Thus, the probability of the second occurring changed based on our knowledge that the first occurred. • It follows, then, that the two events are not independent. • A common error is to treat disjoint events as if they were independent, and apply the Multiplication Rule for independent events—don’t make that mistake.

7. Disjoint vs. Independence • The prerequisite for a required course is that students must have taken either course A or course B. • By the time they are juniors, 57% of the students have taken course A, 21% have had course B, and 15% have done both.

8. Are the events of test A and B disjoint? • Yes, because the outcome of one influences the probability of the other • Yes, because the outcome of one does not influence the probability of the other • No, because there are common outcomes between them • Yes, because there are no common outcomes

9. What percent of the juniors are ineligible for the course? • 57 +21 • 57+21-15 • 100-(57+21) • 100-(57+21-15)

10. Independence • Independence of two events means that the outcome of one event does not influence the probability of the other. • With our new notation for conditional probabilities, we can now formalize this definition: • Events A and B are independent whenever P(B|A) = P(B).(Equivalently, events A and B are independent whenever P(A|B) = P(A).)

11. What is the probability that a junior who has taken course A has also taken course B? • (57+21-15)/(57) • (57+21-15)/(21) • (15)/(57) • (15)/(21)

12. Are the events independent? • No, because the outcome of one influences the probability of the other • Yes, because the outcome of one does not influence the probability of the other • No, because there are common outcomes between them • Yes, because there are no common outcomes

13. Drawing Without Replacement • Sampling without replacement means that once one object is drawn it doesn’t go back into the pool. • We often sample without replacement, which doesn’t matter too much when we are dealing with a large population. • However, when drawing from a small population, we need to take note and adjust probabilities accordingly. • Drawing without replacement is just another instance of working with conditional probabilities.

14. Homework Problem 7 – dealing without replacement • You are dealt a hand of three cards, one at a time. • 4 suits (hearts, clubs, spades, diamond) • 26 red • 26 black • 12 face cards • 4 Aces

15. What’s the probability the first RED card you get is the 3rd card dealt? • ½ + ½ + ½ • ½ * ½ * ½ • 26/52 * 25/51 * 26/50 • 26/52 * 26/52 * 26/52

16. What is the probability that your cards are all hearts? • 13/52 * 12/51 * 11/50 • 26/52 * 25/51 * 24/50 • ¼ * ¼ * ¼ • ½ * ½ * ½

17. What’s the probability that you get no black cards? • ¾ * ¾ * ¾ • ¼ * ¼ * ¼ • 39/52 * 38/51 * 37/50 • 26/52 * 25/51 * 24/50

18. What’s the probability that you have at least one spade? • 1- 13/52 * 12/51 * 11/50 • 1- ¼ * ¼ * ¼ • 1- ¼ • 1- 39/52 * 38/51 * 37/50

19. Battery problem • A junk box in your room contains 15 old batteries, 7 of which are totally dead. • You start picking batteries one at a time and testing them. • Find the probability of each outcome

20. The first 2 you choose are good. • 7/15 * 7/15 • 7/15 * 6/14 • 8/15 * 8/15 • 8/15 * 7/14

21. At least one of the first 4 batteries work. • 8/15 * 8/15 * 8/15 * 8/15 • 1-8/15 * 8/15 * 8/15 * 8/15 • 1- 8/15 * 7/14 * 6/13 * 5/12 • 1- 7/15 * 6/14 * 5/13 * 4/12

22. You have to pick 5 batteries to find one that works. • 7/15 * 6/14 * 5/13 * 4/12 * 8/11 • 7/15 * 6/14 * 5/13 * 5/12 * 4/11 • 1 – 8/15 * 7/14 * 6/13 * 5/12 * 4/11 • 8/15 * 7/14 * 6/13 * 5/12 * 4/11

23. Reversing the conditioning of two events is rarely intuitive. Suppose we want to know P(A|B), but we know only P(A), P(B), and P(B|A). We also know P(AandB), since P(AandB) = P(A) x P(B|A) From this information, we can find P(A|B): Reversing the Conditioning

24. Sobriety checkpoint problem • Police establish a sobriety checkpoint where they detain drivers whom they suspect have been drinking and release those who have not. The police detain 81% of drivers who have been drinking and release 81% of drivers who have not. • Assume that 10% of drivers have been drinking.

25. What’s the probability of any given driver will be detained? • .10*.81+.9*.19 • .10*.19+.9*.81 • .10*.81 • .9*.81

26. What’s the probability that a driver who is detained has actually been drinking? • 25.2 / 10 • 8.1 / 25.2 • 10 / 25.2 • 81 / 25.2

27. What’s the probability that a driver who was released had actually been drinking? • 1.9 / 25.2 • 72.9 / 25.2 • 1.9 / 74.8 • 72.9 / 25.2

28. One more Tree Problem • Dan’s Diner employs three dishwashers. • Al washes 60% of the dishes and breaks only 2% of those he handles. • Betty and Chuck each wash 20% of the dishes, and Betty breaks only 2% of hers, but Chuck breaks 4% of the dishes he washes. • You go to Dan’s for supper one night and hear a dish break at the sink.

29. What’s the probability Chuck broke the dish? • .008/(.008+.004+.012) • .012/(.008+.004+.012) • .008/(.04) • .008/(.96)

30. What’s the probability that Al broke the dish? • .008/(.008+.004+.012) • .012/(.008+.004+.012) • .008/(.04) • .008/(.96)

31. Upcoming In Class • Homework 5 due Sunday 2/17 • Quiz 3 next Thursday – February 21st • (HW4 and HW5) • Exam 1 – March 7th