5-Minute Check on Section 6-3a

1. 5-Minute Check on Section 6-3a Fifty animals are to be used in a stress study: 3 male and 6 female dogs, 9 male and 7 female cats, 5 male and 9 female monkeys, 6 male and 5 female rats. Find the probability of choosing: P(female animal) P(male or a cat) P(female and a monkey) P(animal other than a rat) P(cat or a dog) = females / total = 27 / 50 = 0.54 or 54% = P(male) + p(cat) – P(male cat) = 0.46 + 0.32 – 0.18 = 0.60 or 60% = P(female monkey) = 9 / 50 = 0.18 or 18% = 1 – P(rat) = 1 – 11 / 50 = 39 / 50 = 0.78 or 78% = P(cat) + p(dog) – P(cat-dog) = 0.32 + 0.18 – 0.00 = 0.50 or 50% Click the mouse button or press the Space Bar to display the answers.

2. Lesson 6 – 3b General Probability Rules

3. Knowledge Objectives • Define what is meant by a joint event and joint probability • Explain what is meant by the conditional probability P(A | B) • State the general multiplication rule for any two events • Explain what is meant by Bayes’s rule.

4. Construction Objectives • State the addition rule for disjoint events • State the general addition rule for union of two events • Given any two events A and B, compute P(AB) • Given two events, compute their joint probability • Use the general multiplication rule to define P(B | A) • Define independent events in terms of a conditional probability

5. Vocabulary • Personal Probabilities – reflect someone’s assessment (guess) of chance • Joint Event – simultaneous occurrence of two events • Joint Probability – probability of a joint event • Conditional Probabilities – probability of an event given that another event has occurred

6. General Multiplication Rule The probability that two events A and B both occur is P(A and B) = P(A  B) = P(A) ∙ P(B | A) where P(B | A) is a conditional probability read as the probability of B given that A has occurred

7. Conditional Probability Rule If A and B are any two events, then P(A and B) N(A and B) P(B | A) = ----------------- = ---------------- P(A) N(A) N is the number of outcomes

8. Independence in Terms of Conditional Probability Two events A and B are independent if P(B | A) = P(B) Example: P(A = Rolling a six on a single die) = 1/6 P(B = Rolling a six on a second roll) = 1/6 no matter what was rolled on the first roll!! So probability of rolling a 6 on the second roll, given you rolled a six on the first is still 1/6P(B | A) = P(B) so A and B are independent

9. Contingency Tables • What is the probability of left-handed given that it is a male? • What is the probability of female given that they were right-handed? • What is the probability of being left-handed? P(LH | M) = 12/60 = 0.20 P(F| RH) = 42/90 = 0.467 P(LH) = 20/110 = 0.182

10. Tree Diagram 0.8 Right-handed 0.44 How do we get the probabilities on the far right from the table? Male 0.55 0.2 Left-handed 0.11 Sex Right-handed 0.84 0.378 0.45 Female Left-handed 0.16 0.072 Right-handed males: 0.44 = 48/110 Left-handed females: 0.072 = 8/110

11. Example 1 A construction firm has bid on two different contracts. Let B1 be the event that the first bid is successful and B2, that the second bid is successful. Suppose that P(B1) = .4, P(B2) = .6 and that the bids are independent. What is the probability that: a) both bids are successful? b) neither bid is successful? c) is successful in at least one of the bids? Independent  P(B1) • P(B2) = 0.4 • 0.6 = 0.24 Independent  (1- P(B1)) • (1 - P(B2)) = 0.6 • 0.4 = 0.24 3 possible outcomes  (1- P(a)- P(b)) = 1 – 0.24 – 0.24 = 0.52 or P(B1) • (1 – P(B2)) + (1 – P(B1)) • P(B2) = 0.4 • 0.4 + 0.6 • 0.6 = 0.52

12. Example 2 Given that P(A) = .3 , P(B) = .6, and P(B|A) = .4 find: a) P(A and B) b) P(A or B) c) P(A|B) P(A and B) P(B|A) = ----------------- so P(A and B) = P(B|A)•P(A) P(A) P(A and B) = 0.4 • 0.3 = 0.12 P(A or B) = P(A) + P(B) – P(A and B) = 0.3 + 0.4 – 0.12 = 0.58 P(A and B) 0.12 P(A|B) = ----------------- = -------- = 0.2 P(B) 0.6

13. Example 3 Given P(A | B) = 0.55 and P(A or B) = 0.64 and P(B) = 0.3. Find P(A). P(A and B) P(A|B) = ----------------- so P(A and B) = P(A|B)•P(B) P(B) P(A and B) = 0.55 • 0.3 = 0.165 P(A or B) = P(A) + P(B) – P(A and B) P(A) = P(A or B) – P(B) + P(A and B) = 0.64 – 0.3 + 0.165 = 0.505

14. Example 4 If 60% of a department store’s customers are female and 75% of the female customers have a store charge card, what is the probability that a customer selected at random is female and had a store charge card? Let A = female customer and let B = customer has a store charge card P(A and B) P(B|A) = ----------------- so P(A and B) = P(B|A)•P(A) P(A) P(A and B) = 0.75 • 0.6 = 0.45

15. Example 5 Suppose 5% of a box of 100 light blubs are defective. If a store owner tests two light bulbs from the shipment and will accept the shipment only if both work. What is the probability that the owner rejects the shipment? P(reject) = P(at least one failure) = 1 – P(no failures) = 1 – P(1st not defective) • P(2nd not defective | 1st not defective) = 1 – (95/100) • (94/99) = 1 – 0.9020 = 0.098 or 9.8% of the time

16. Question to Ponder • Dan can hit the bulls eye ½ of the time • Daren can hit the bulls eye ⅓ of the time • Duane can hit the bulls eye ¼ of the time Given that someone hits the bulls eye, what is the probability that it is Dan? Out of 36 throws, 13 hit the target. Dan had 6 of them, so P(Dan | bulls eye) = 6/13 = 0.462

17. Summary and Homework • Summary • Two events are independent if • P(A|B) = P(A) and P(B|A) = P(B) • P(at least one) = 1 – P(none) (complement rule) • Homework • Day Two: 6.72, 73, 76, 81, 82