ability Notes Definitions: Sample Space: sample space is a set or collection of possible outcomes. Flipping a Coin: {Head, Tail} Rolling Two Die: {,,,, 6, 7, 8, 9, 0,, } Outcome: n outcome is an element of a sample space. Head roll of Event: n event is any subset of a sample space. Flipping a Head Rolling and even number, {,, 6, 8, 0, } Compliment: The compliment of an event is the set of all outcomes from the sample space not in the event. The compliment of getting a Head is getting a Tail. The compliment of getting an even roll is getting an odd roll. ability: For each outcome in a sample space there is a probability, which is the likelihood of choosing that outcome at random from the sample space. The probability of a Head is. The probability of rolling a is. The probability of an event is the likelihood of choosing an outcome from that event. The probability of rolling a prime number is + + + 6 + = ability Space: probability space is a sample space together with a set of probabilities. C H T R 6 7 8 9 0 X 6 7 8 9 0 nd: If and are events in a sample space, the probability of and is the sum of probabilities of all the outcomes in both and. P ( and ) = 6 s i and The probability that a roll is even and prime: probability of a. The probability that a roll is odd and greater than : probability of a 7, 9, or. Or: If and are events in a sample space, the probability of or is the sum of the probabilities of all the outcomes in either or. P ( or ) = s i or The probability that a roll is odd or less than 7: probability of a,,,, 6, 7, 9, or.
The probability that a roll is divisible by or : probability of, 8,,, 6, or 9. Disjoint: Two events are disjoint if they contain none of the same outcomes. The event of rolling an even and the event of rolling an odd are disjoint. Independent: Two events are independent if knowing the outcome of one does not effect the probability of the other. The event that your first flip is a head and the event that your second flip is a tail are independent. The event that one die is a and the event that the other die is a are independent. Conditional abilities: If and are events in a sample space, the conditional probability of given is the probability that will occur if we know the outcome of. The probability that the second card dealt is red given that the first card dealt is black. P (C = Red C = lack) = 6 The probability that a roll is prime given that the roll is odd. P (R = prime R = odd) = 8 True Odds: If is an event in a sample space with probability P (), then the true odds of are written X : Y against (read X to Y against ), where P () = Y (X+Y ) and ( P ()) = X (X+Y ). The proability of rolling a 7 or an on two dice is /9 so the true odds of winning on the first roll in craps is 7: against. The probability of being dealt two cards totaling 9 in accarat is 6/6, so the odds of winning on a deal of 9 is 96:6 against. The odds of successfully navigating an asteroid field are 70 : against, so the probability that you could do so is /7 and the probability you couldn t is 70/7. House Odds: The house odds of an event are X : Y if a bet of $Y pays a return of $X if you win. The house odds of winning on a single number bet in roulette are :, so on a one dollar bet you win $, plus you get your one dollar back. House dvantage (Expected Value): If the house odds of an event are X : Y and the probability of is P () then the house advantage or expected value is X P () + ( Y ) ( P ()). The house odds of winning on a single number bet in roulette are : and the probability of winning that bet is /8, so the house advantage is 7 + ( ) 8 8 = 0.06 or.6%. 8 This means that on average, over time, the house wins for every dollar bet, for a ten dollar bet, or $.6 for a hundred dollar bet.
Rules: Venn Diagram: Let S be a sample space, s i be outcomes in S, and be events in S, and and c be the compliments of and in S. The notation P ( ) is read, The probability of given.. 0 P ( and ) P ( and ) P (). P (S) = i =. P () = s i. P () + P ( ) = c P ( and c ) P ( and c ) P ( c ). P ( or ) = P () + P () P ( and ) (General ddition Rule) 6. If and are disjoint, then P ( or ) = P () + P () and P ( and ) = 0. 7. P ( and ) = P ()P ( ) (General Multiplication Rule) 8. If and are independent, then P ( ) = P () and P ( and ) = P ()P (). 9. P ( and ) + P ( and c ) = P () 0. P ( or ) = P () + P ( ) =. P ( and ) = 0. P ( ) = P ()P ( ) P ()P ( )+P ( )P ( ) (ayes Theorem) Example: P () P ( ) Let be the event that I roll an odd number, and let be the event that I roll a prime number. First find the probability of each event. Roll 6 7 8 9 0 So the probability of is: P () = P (,, 7, 9, or ) = P () + p() + P (7) + P (9) + P () = 8, and the probability of is: 6 P () = P (,,, 7, or ) =.
Next, find the probability of and : P ( and ) = P (,, 7, or ) =. Finally we can use the rules to fill an the entire Venn Diagram. P ( ) = P () = 8 P ( c ) = P () = P ( and c ) = P () P ( and ) = P ( and ) = P () P ( and ) = P ( and c ) = P ( ) P ( and ) = 7 Exercise: Use the probability rules and definitions in order to fill in a Venn Diagram with the given information. P ( or ) = P () = and are disjoint P ( and ) P ( and ) P () P ( and c ) P ( and c ) c P ( c ) c 7 P () P ( ) 8 8
Decision Tree: P ( ) P ( and ) 6 6 P (c ) c P ( and ) P ( and c ) P () P (c ) c P ( and c ) P ( ) P ( and ) P ( ) P ( ) P (c ) c P ( and ) P ( and c ) P (c ) c P ( and c ) Example: Make a decision tree were event is the first card dealt is red, and event is the second card dealt is black. Exercise: Fill in a decision tree using the given information: P ( and ) = 8 First let s calculate a few probabilities: P ( c ) = P () = 6 P ( ) = 6 P ( ) = Now an observation: P ( ) + P ( c ) = P ( and ) P () + P ( and c ) P () =, P ( ) = 6 P () P ( ) P ( ) P (c ) P ( ) c P ( and ) P ( and c ) P ( and ) therefore, we can fill in the rest of the probabilities with what we already have. P (c ) c P ( and c )