Lesson 6 – 3b General Probability Rules Taken from And modified slightly

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.

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

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

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

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

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 1/6 P(B | A) = P(B) so A and B are independent

Contingency Tables MaleFemaleTotal Right handed Left handed12820 Total What is the probability of left-handed given that it is a male? 2.What is the probability of female given that they were right-handed? 3.What is the probability of being left-handed? P(LH | M) = 12/60 = 0.20 P(F| RH) = 42/90 = P(LH) = 20/110 = 0.182

Tree Diagram Sex Female Male Left-handed Right-handed Left-handed Right-handed

Example 1 A construction firm has bid on two different contracts. Let B 1 be the event that the first bid is successful and B 2, that the second bid is successful. Suppose that P(B 1 ) =.4, P(B 2 ) =.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(B 1 ) P(B 2 ) = = 0.24 Independent  (1- P(B 1 )) (1 - P(B 2 )) = = – P(neither bid is successful) = =.76

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) = = 0.12 P(A and B) 0.12 P(A|B) = = = 0.2 P(B) 0.6 P(A or B) = P(A) + P(B) – P(A and B) = – 0.12 = 0.78

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) = = P(A or B) = P(A) + P(B) – P(A and B) Rearrange Formula P(A) = P(A or B) – P(B) + P(A and B) = 0.64 – = 0.505

Example 4 If 60% of a department store’s customers are female and 75% of the female customers have a store charge card (assuming these events are independent), 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(A)P(B) P(A and B) = = 0.45

Example 5 - Independent 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? (Assume each light bulb is independent) P(accept) = P(1 st not defective AND 2 nd not defective) = =.9025 P(reject) = 1 - P(accept) = 1 –.9025 =.0975 or 9.75% of the time

Example 5 - Dependent 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? (Assume each light bulb is dependent) P(1 st not defective) = 95/100 P(2 nd not defective | 1 st not defective) = 94/99 P(accept) = P( both are not defective) = ( 95/100) (94/99) =.9020 P(reject) = 1 - P(accept) = 1 –.9020 =.098 or 9.8% of the time

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? HitMiss Dan (p=1/2)66 Daren (p=1/3)48 Duane (p=1/4)39 Out of 36 throws, 13 hit the target. Dan had 6 of them, so P(Dan | bulls eye) = 6/13 = 0.462

Summary and Homework Summary Homework –Day Two: 6.72, 73, 76, 81, 82