Advanced Placement Statistics Ch. 6.3: General Probability Rules & Multistage Probability EQ: How do you calculate joint and conditional probability?

For any event A and its complement, P(Ac) = 1 – P(A). Important Recall Complete each Statement For any event A and its complement, P(Ac) = 1 – P(A). When two events, A and B, are independent, then P(A Ç B) = P(A)  P(B)

If two events, A and B, are disjoint, then P(A È B) = P(A) + P(B). Given two events, A and B, then P(A È B) = P(A) + P(B) – P(A  B).

Contingency Tables --- two-way tables giving frequency for categorical variables Joint Probabilities --- simultaneous occurrence of 2 events

P(heard ad and bought product) = 0.15 Complete the contingency table. The following information was given about the success of an ad campaign. P(heard ad) = 0.35 P(bought product) = 0.23 P(heard ad and bought product) = 0.15 Complete the contingency table. 0.15 0.08 0.23 0.20 0.57 0.77 0.65 1.00 0.35

Assignment: p. 440 #65 - 68 Find P(did not hear ad) = _______ 0.65 Find P(did not hear ad) = _______ 2. Find P(did not buy product) = _________ 0.77 Assignment: p. 440 #65 - 68

Conditional Probability --- probability of one event on the condition we know another event

** If P(B | A) = P(B| A’) = P(B) what can we say about events A and B? Event A and event B are independent events. The probability of event B is not different if event A has or has not occurred.

Events A and B are independent, so B does not have to occur on the CONDITION event A occurred. This is the intersection of events “A and B” occurring.

Recall: Not Independent Events: If A and B are not independent events, the probability of B happening will depend upon whether A has happened or not. We therefore have to introduce conditional probabilities in the tree diagram (as shown below):

Events A and B are not independent, so B does occur on the CONDITION event A occurred. This is the intersection of events “A and B given A” occurring.

Just ALGEBRA!!

Create a tree diagram. Use T as read The Times and M as read The Mail Create a tree diagram. Use T as read The Times and M as read The Mail. Use C as completed the crossword puzzle and C’ as not completed the crossword puzzle.

RECALL: Multiplication Rule for Independent Events: If events A and B are independent events,   then we can say P(A and B) = P(A) · P(B). The Converse Statement of This Rule States:   If P(A) · P(B) = P(A and B), then we can assume A and B are independent events. USE THIS STATEMENT TO JUSTIFY INDEPENDENCE!!!

In class: Conditional Probability Worksheet #1 - 15

Earlier you found the P(married | age 18 to 24) = 0. 241 Earlier you found the P(married | age 18 to 24) = 0.241. Complete the sentence: 24.1 % is the proportion of women who are __________________________ among those women who are _________________________. married age 18 to 24 In #13 you found P(age 18 to 24 | married). Write a sentence in the form given in #14 that describes the meaning of this result. 5.2% is the proportion of women who are age 18 to 24 among those women who are married.

  Slim is a professional poker player. At the moment he wishes very much to draw two diamonds in a row. As he sits at the table looking at his hand and at the upturned cards on the table, Slim sees 11 cards. Of these, 4 are diamonds. The full deck contains 13 diamonds among the 52 cards in the deck, so 9 of the 41 unseen cards are diamonds. The deck was shuffled, so each card Slim draws is equally likely to be any of the cards that he has not seen. A = 1st draw is a diamond B = 2nd card is a diamond 1. What is the probability that Slim draws a diamond on his first card? P(A) = 9/41

  Slim is a professional poker player. At the moment he wishes very much to draw two diamonds in a row. As he sits at the table looking at his hand and at the upturned cards on the table, Slim sees 11 cards. Of these, 4 are diamonds. The full deck contains 13 diamonds among the 52 cards in the deck, so 9 of the 41 unseen cards are diamonds. The deck was shuffled, so each card Slim draws is equally likely to be any of the cards that he has not seen. A = 1st draw is a diamond B = 2nd card is a diamond 1. What is the probability that Slim draws a diamond on his first card? P(A  B) = P(A) • P(B|A) = 9/41 • 8/40 = .044 P(A  B’) = P(A) • P(B’|A) = 9/41 • 32/40 = .1756 P(A’  B) = P(A’) • P(B|A’) = 32/41 • 9/40 = .1756 P(A’  B’) = P(A’) • P(B’|A’) = 32/41 • 31/40 = .6049

P(diamond and diamond) = P(A  B) = P(A) • P(B | A) = Write the conditional probability statement that meets Slim’s wish. Calculate this probability using the conditional probability multiplication formula.   Would you assume these events to be independent? Justify your reason mathematically. 0.044 0.1756 0.6049 P(diamond and diamond) = P(A  B) = P(A) • P(B | A) = If A and B are independent then what must be true? Therefore A and B are not independent.

Only 5% of male high school basketball, baseball, and football players go on to play at the college level. Of these, only 1.7% play major league professional sports. Only 0.01% of professional athletes did not compete in college. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years.   4. Construct a tree diagram to represent the following events. Use letter notation and percentages to define the parts of the tree. Let A = competes in college Let B = competes professionally Let C = pro career longer than 3 years Intersection probability Conditional probability P(A  B) = P(A) • P(B|A) = .05 • .017 = .00085 P(A  B’) = P(A) • P(B’|A) = .05 • .983 = .04915 P(A’  B) = P(A’) • P(B|A’) = .95 • .0001 = .000095 P(A’  B’) = P(A’) • P(B’|A’) = .95 • .9999 = .949905

Only 5% of male high school basketball, baseball, and football players go on to play at the college level. Of these, only 1.7% play major league professional sports. Only 0.01% of professional athletes did not compete in college. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years.   4. Construct a tree diagram to represent the following events. Use letter notation and percentages to define the parts of the tree. Let A = competes in college Let B = competes professionally Let C = pro career longer than 3 years

Find the following probabilities. Write the verbal statements for each Find the following probabilities. Write the verbal statements for each. Show your work under each problem.   P(A) = P( _____________________) =  P(B | A) =P (__________________________________________) = P(A and B) = P(________________________________________________) =

P(B and Ac) = P( ________________________________________________________) =   P(Bc | A) = P(___________________________________________________________) =

P(B | Ac) = =P(_____________________________________________)=   P(Bc | Ac) = P(_____________________________________________________) =  P(C | A and B) = P( ______________________________________________________) =

P(B) = P(A  B) • P (A’  B) = 0.00085 + 0.000095 = 0.000945 P(B | Ac) = =P(_____________________________________________)=   P(Bc | Ac) = P(_____________________________________________________) = P(C | A and B) = P( ______________________________________________________) = What proportion athletes compete professionally? P(B) = P(A  B) • P (A’  B) = 0.00085 + 0.000095 = 0.000945

7. What is the probability that a high school athlete competes in college and then plays a professional sport for at least 3 years? That is P(A and B and C).   P(A and B and C) = P(A) ·P(B|A) ·P(C | A ∩ B) = ________________________________ 0.05 • 0.017 • 0.4 = 0.00034

ASSIGNMENT: p. 452 #71 – 76, 79 - 84