1. The probability of an event based on the fact that some other event has occurred, will occur, or is occurring. P(B/A) = Conditional Probability The probability of event B occurring given that event A has occurred is usually stated as “the conditional probability of B, given A; P(B/A) 11.3 – Conditional Probability – Events Involving “And”

2. Example: A number from the sample space S = {2, 3, 4, 5, 6, 7, 8, 9} is randomly selected. Given the defined events A and B, A: selected number is odd, and B: selected number is a multiple of 3 find the following probabilities. a) B = {3, 6, 9} Conditional Probability b) P(A and B) = a) P(B) b) P(A and B) c) P(B/A) 11.3 – Conditional Probability – Events Involving “And” P(B) =3/8 P({3, 5, 7, 9}  {3, 6, 9}) = P({3, 9}) =2/8 =1/4 c) Probability of B given A has occurred: P(B/A) = P(A) P(A and B) 4/8 1/4 1/2 = =

3. Example: Given a family with two children, find the probability that both are boys, given that at least one is a boy. P({gb, bg, bb}) = Conditional Probability P(A and B) = A = at least one boy 11.3 – Conditional Probability – Events Involving “And” P(A) = 3/4 P({gb, bg, bb}  {bb}) = P({bb}) = = 1/3 1/4 Conditional ProbabilityP(B/A) = 3/4 1/4 B = both are boys S= {gg, gb, bg, bb} A = {gb, bg, bb} B = {bb} =

4. Two events are Independent if the occurrence of one of them has no effect on the probability of the other. P(B/A) = P(B) Independent Events or 11.3 – Conditional Probability – Events Involving “And” P(A/B) = P(A)

5. Example: A single card is randomly selected from a standard 52-card deck. Given the defined events A and B, A: the selected card is an ace, B: the selected card is red, find the following probabilities. a) P(B) = Independent Events b) P(A and B) = a) P(B) b) P(A and B) c) P(B/A) 11.3 – Conditional Probability – Events Involving “And” = 1/2 P({Ah, Ad, Ac, As}  {all red}) = P({Ah, Ad}) = 2/52 Events A and B are independent as P(B) = P(B/A). c) P(B/A) = P(A) P(A and B) 4/52 2/52 1/2 = = 52 26

6. If A and B are any two events then P(A and B) = P(A)  P(B/A) Multiplication Rule of Probability - Events Involving “And” 11.3 – Conditional Probability – Events Involving “And” P(A and B) = P(A)  P(B) If A and B are independent events then A jar contains 4 red marbles, 3 blue marbles, and 2 yellow marbles. What is the probability that a red marble is selected and then a blue one without replacement? P(Red and Blue) = P(Red)  P(Blue/Red) = 4/9  3/8 = 12/72 = 1/8= 0.1667 Example:

7. Multiplication Rule of Probability - Events Involving “And” 11.3 – Conditional Probability – Events Involving “And” A jar contains 4 red marbles, 3 blue marbles, and 2 yellow marbles. What is the probability that a red marble is selected and then a blue one with replacement? P(Red and Blue) = P(Red)  P(Blue) = 4/9  3/9 = 12/81 = 4/27= 0.148 Example:

8. 11.5 – Expected Value The Expected Value of x is the sum of the products of the values of x and their corresponding probabilities. E(x) = x 1 P(x 1 ) + x 2 P(x 2 ) + x 3 P(x 3 ) + … + x n P(x n ) The expected value is a calculation that serves as the best prediction of a value. It is the probability-weighted average of all possible outcomes. The expected value of a possible future event assists in making mathematically sound decisions. It is often used when making investments, determining a price for numerous services, prioritizing events, and in calculating Return on Investment.

9. 11.5 – Expected Value Example: A third grade class was surveyed regarding the number of hours that they played electronic games each day. The probability distribution is given in the table below: # of Hours (x)Probability P(x) 00.3 10.4 20.2 30.1 Calculate the Expected Value of the quantity of time that a third grader spends each day playing electronic games.

10. 11.5 – Expected Value E(x) = x 1 P(x 1 ) + x 2 P(x 2 ) + x 3 P(x 3 ) + … + x n P(x n ) # of Hours (x)Probability P(x) 00.3 10.4 20.2 30.1 Expected value, E(x) = 0 (0.3) + 1 (0.4) + 2 (0.2) + 3 (0.1) Expected value, E(x) = 0 + 0.4 + 0.4 + 0.3 Expected value, E(x) = 1.1 hours

11. 11.5 – Expected Value Example: Find the expected number of boys for a three-child family. Assume girls and boys are equally likely. # of BoysProbabilityProduct xP(x)x P(x) 01/8 13/8 2 31/8 Expected value, E(x) = 0 + 3/8 + 6/8 + 3/8 Expected value, E(x) = 12/8 Expected value, E(x) = 3/2 Expected value, E(x) = 1.5 boys 0 3/8 6/8 3/8 bbb bbg bgb bgg gbb gbg ggb ggg

12. 11.5 – Expected Value Example: Finding Expected Winnings Win $7 by rolling a 6 on a single die, Win $1 by rolling any other number. What are the expected net winnings for the game? NumberPayoffNetP(x)x P(x) 1, 2, 3, 4, 5 6 A player pays $3 to play the following game:

13. 11.5 – Expected Value Example: A player pays $3 to play the following game: Expected value, E(x) = – $10/6 + $4/6 Expected value, E(x) = – $6/6 Expected value, E(x) = – $1 Win $7 by rolling a 6 on a single die, Win $1 by rolling any other number. What are the expected net winnings for the game? NumberPayoffNetP(x)x P(x) 1, 2, 3, 4, 5$1- $25/6- $10/6 6$7$41/6$4/6 Finding Expected Winnings

14. 11.5 – Expected Value Previous Example: Finding Expected Winnings A player pays $3 to play the following game: A fair game is one in which the net winnings are zero. Win $7 by rolling a 6 on a single die, Win $1 by rolling any other number. Is this a fair game? An unfair game against the player has negative expected winnings. An unfair game in favor of the player has positive expected winnings. It is an unfair game against the player as it has negative expected winnings (–$1). NO

15. 11.5 – Expected Value Example: Expected Investment Profits E(x) = -400 (0.2) + 800 (0.5) + 1300 (.3) Mark intends to invest $6,000 in one of two companies. His research is presented in the tables below: What are the expected profits (or loses) for each company? Company ABC Profit/Loss (x)Probability P(x) - $4000.2 $8000.5 $13000.3 Company PDQ Profit/Loss (x)Probability P(x) $6000.8 $10000.2 E(x) = -80 + 400 + 390 E(x) = $710 E(x) = 600 (0.8) + 1000 (0.2) E(x) = 480 + 200 E(x) = $680 Profit