1. Lesson 5 - 4 Conditional Probability and the General Multiplication Rule

2. Objectives Compute conditional probabilities Compute probabilities using the General Multiplication Rule

3. Vocabulary Conditional Probability – probability that an event F occurs, given that event E has occurred. P(E|F) is the symbols for the above and is read probability of F given E has occurred.

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

5. General Multiplication Rule The probability that two events E and F both occur is P(E and F) = P(E) ∙ P(E|F)

6. Independence in Terms of Conditional Probability Two events E and F are independent if P(E|F) = P(F) Example: P(E=Rolling a six on a single die) = 1/6 P(F=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(E|F) = P(F) so E and F are independent

7. Contingency Tables MaleFemaleTotal Right handed384280 Left handed12820 Total50 100 1.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) = 12/50 = 0.24 P(F| RH) = 42/80 = 0.525 P(LH) = 20/100 = 0.20

8. 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 ) = 0.4 0.6 = 0.24 Independent  (1- P(B 1 )) (1 - P(B 2 )) = 0.6 0.4 = 0.24 3 possible outcomes  (1- P(a)- P(b)) = 1 – 0.24 – 0.24 = 0.52 or P(B 1 ) (1 – P(B 2 )) + (1 – P(B 1 )) P(B 2 ) = 0.4 0.4 + 0.6 0.6 = 0.52

9. 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 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) = 0.3 + 0.4 – 0.12 = 0.58

10. 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

11. 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

12. 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(1 st not defective) P(2 nd not defective | 1 st not defective) = 1 – (95/100) (94/99( = 1 – 0.9020 = 0.098 or 9.8% of the time

13. Summary and Homework Summary –Conditional probabilities P(F|E) represent the chance that F occurs, given that E occurs also –The General Multiplication Rule applies to “and” problems for all events and involves conditional probabilities Homework –pg 289 - 292: 3, 5, 7, 9, 12, 19, 27, 31, 32