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1 Probability Rules Probability II Chapter 4 1. The probability P(A) for any event A is 0 P(A) 1.. If S is the sample space in a probability model, then P(S)=1. 3. For any event A, P(A does not occur) = 1- P(A). 4. If A and B are disjoint events, P(A or B)=P(A)+P(B). 5. For any two events, P(A or B) = P(A)+P(B)-P(A and B). More rules: P(A and B) Another situation of frequent interest, finding P(A and B), the probability that both events A and B occur. P(A and B)= P(event A occurs and event B occurs) As for the Addition rule, we have two versions for P(A and B). Independent events Two events are independent if the probability that one event occurs on any given trial of an experiment is not influenced in any way by the occurrence of the other event. : toss a coin twice Event A: first toss is a head (H) Event B: second toss is a tail (T) Events A and B are independent. The outcome of the first toss cannot influence the outcome of the second toss. Imagine coins spread out so that half were heads up, and half were tails up. Pick a coin at random. The probability that it is headsup is 0.5. But, if you don t put it back, the probability of picking up another heads-up coin is now less than 0.5. Without replacement, successive trials are not independent. In this example, the trials are independent only when you put the coin back ( sampling with replacement ) each time. A woman's pocket contains quarters and nickels. She randomly extracts one of the coins and, after looking at it, replaces it before picking a second coin. Let Q1 be the event that the first coin is a quarter and Q be the event that the second coin is a quarter. Are Q1 and Q independent events? YES! Why? Since the first coin that was selected is replaced, whether Q1 occurred (i.e., whether the first coin was a quarter) has no effect on the probability that the second coin is a quarter, P(Q). In either case (whether Q1 occurred or not), when we come to select the second coin, we have in our pocket: 1

2 More s EXAMPLE: Two people are selected at random from all living humans. Let B1 be the event that the first person has blue eyes and B be the event that the second person has blue eyes. In this case, since the two were chosen at random, whether the first person has blue eyes has no effect on the likelihood of choosing another blue eyed person, and therefore B1 and B are independent. On the other hand... EXAMPLE: A family has 4 children, two of whom are selected at random. Let B1 be the event that the first child has blue eyes, and B be the event that the second chosen child has blue eyes. In this case B1 and B are not independent since we know that eye color is hereditary, so whether the first child is blue-eyed will increase or decrease the chances that the second child has blue eyes, respectively. Decide whether the events are independent or dependent Event A: a salmon swims successfully through a dam Event B: another salmon swims successfully through the same dam Event A: parking beside a fire hydrant on Tuesday Event B: getting a parking ticket on the same Tuesday The Multiplication Rule for independent events If A and B are independent, P(A and B) = P(A) P(B) The Multiplication Rule for independent events: Roll two dice. Event A: roll a 6 on a red die Event B: roll a 5 on a blue die. What is the probability that if you roll both at the same time, you will roll a 6 on the red die and a 5 on the blue die? Since the two dice are independent, P(A and B)=P(A) P(B)=1/6 1/6 = 1/36 Recall the blood type example: Two people are selected at random from all living humans. What is the probability that both have blood type O? Let O1= "the first has blood type O" and O= "the second has blood type O" We need to find P(O1 and O) Since the two were chosen at random, the blood type of one has no effect on the blood type of the other. Therefore, O1 and O are independent and we may apply Rule 5: P(O1 and O) = P(O1)*P(O) =.44*.44= Probability Rules 1. The probability P(A) for any event A is 0 P(A) 1.. If S is the sample space in a probability model, then P(S)=1. 3. For any event A, P(A does not occur) = 1- P(A). 4. If A and B are disjoint events, P(A or B)=P(A)+P(B). 5. General Addition Rule: For any two events, P(A or B) = P(A)+P(B)-P(A and B). 6. If A and B are independent, P(A and B) = P(A) P(B)

3 One more rule We need a general rule for P(A and B). For this, first we need to learn Conditional probability. Notation: P(B A). This means the probability of event B, given that event A has occurred. Event A represents the information that is given. All the students in a certain high school were surveyed, then classified according to gender and whether they had either of their ears pierced: Suppose a student is selected at random from the school. Let M and not M denote the events of being male and female, respectively, and E and not E denote the events of having ears pierced or not, respectively. M not M E not E What is the probability that the student has one or both ears pierced? Since a student is chosen at random from the group of 500 students out of which 34 are pierced, P(E)=34/500=.648 What is the probability that the student is a male? Since a student is chosen at random from the group of 500 students out of which 180 are males, P(M)=180/500=.36 What is the probability that the student is male and has ear(s) pierced? Since a student is chosen at random from the group of 500 students out of which 36 are males and have their ear(s) pierced, P(M and E)=36/500=.07 Conditional Probability Now something new: Given that the student that was chosen is a male, what is the probability that he has one or both ears pierced? We will write "the probability of having one or both ears pierced (E), given that a student is male (M)" as P(E M). We call this probability the conditional probability of having one or both ears pierced, given that a student is male: it assesses the probability of having pierced ears under the condition of being male. The total number of possible outcomes is no longer 500, but has changed to 180. Out of those 180 males, 36 have ear(s) pierced, and thus: P(E M)=36/180=0.0. 3

4 General formula P( Aand B) P( B A) = P( A) The above formula holds as long as P(A)>0 since we cannot divide by 0. In other words, we should not seek the probability of an event given that an impossible event has occurred. On the "Information for the Patient" label of a certain antidepressant it is claimed that based on some clinical trials, there is a 14% chance of experiencing sleeping problems known as insomnia (denote this event by I), there is a 6% chance of experiencing headache (denote this event by H), and there is a 5% chance of experiencing both side effects (I and H). Thus, P(I)=0.14 P(H)=0.6 P(I and H)=0.05 P(I)=0.14 P(H)=0.6 P(I and H)=0.05 Important!!! (a) Suppose that the patient experiences insomnia; what is the probability that the patient will also experience headache? Since we know (or it is given) that the patient experienced insomnia, we are looking for P(H I). According to the definition of conditional probability: P(H I)=P(H and I)/P(I)=.05/.14=.357. (b) Suppose the drug induces headache in a patient; what is the probability that it also induces insomnia? Here, we are given that the patient experienced headache, so we are looking for P(I H). P(I H)=P(I and H)/P(H)=.05/.6=.193. In general, P( A B) P( B A) Independence Testing for Independence Recall: two events A and B are independent if one event occurring does not affect the probability that the other event occurs. Now that we've introduced conditional probability, we can formalize the definition of independence of events and develop four simple ways to check whether two events are independent or not. Two events A and B are independent if P( B A) = P( B) or P( A and B) = P( A) P( B) Two events A and B are dependent if P( B A) P( B) or P( A and B) P( A) P( B) 4

5 Recall the side effects example. there is a 14% chance of experiencing sleeping problems known as insomnia (I), there is a 6% chance of experiencing headache (H), and there is a 5% chance of experiencing both side effects (I and H). Thus, P(I)=0.14 P(H)=0.6 P(I and H)=0.05 Are the two side effects independent of each other? To check whether the two side effects are independent, let's compare P(H I) and P(H). In the previous part of this lecture, we found that P(H I)=P(H and I)/P(I)=.05/.14=.357 while P(H)=.6 P(H I) P(H) Knowing that a patient experienced insomnia increases the likelihood that he/she will also experience headache from.6 to.357. The conclusion, therefore is that the two side effects are not independent, they are dependent. General Multiplication Rule Comments P(A and B)=P(A)*P(B A) For independent events A and B, we had the rule P(A and B)=P(A)*P(B). Now, for events A and B that may be dependent, to find the probability of both, we multiply the probability of A by the conditional probability of B, taking into account that A has occurred. Thus, our general multiplication rule is stated as follows: Rule 7: The General Multiplication Rule: For any two events A and B, P(A and B)=P(A)*P(B A) This rule is general in the sense that if A and B happen to be independent, then P(B A)=P(B) is, and we're back to Rule 5 - the Multiplication Rule for Independent Events: P(A and B)=P(A)*P(B). Recall the definition of conditional probability: P(B A)=P(A and B)/P(A). Let's isolate P(A and B) by multiplying both sides of the equation by P(A), and we get: P(A and B)=P(A)*P(B A). That's it...this is the General Multiplication Rule. Probability Rules Counting 1. The probability P(A) for any event A is 0 P(A) 1.. If S is the sample space in a probability model, then P(S)=1. 3. For any event A, P(A does not occur) = 1- P(A). 4. If A and B are disjoint events, P(A or B)=P(A)+P(B). 5. General Addition Rule: For any two events, P(A or B) = P(A)+P(B)-P(A and B). 6. If A and B are independent, P(A and B) = P(A) P(B) 7. General Multiplication Rule: P( A and B) = P(A) P(B A) Fundamental Counting Rule Factorial Rule Permutation (when all items are different) Permutation (when some items are like) Combination 5

6 Fundamental Counting Rule For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways. : Suppose you have 3 shirts (call them A, B, and C), and 4 pairs of pants (call them w, x, y, and z). Then you have m = 3, n = 4: 3 4 = 1 possible outfits: Aw, Ax, Ay, Az, Bw, Bx, By, Bz,Cw, Cx, Cy, Cz Notation The factorial symbol! Denotes the product of decreasing positive whole numbers. For example, 6! = = 70 By special definition, 0! = 1. TI-83, 84: MATH PRB 4:! First enter 6, then! Factorial Rule A collection of n different items can be arranged in order n! different ways : Suppose you have four candles you wish to arrange from left to right on your dinner table. The four candles are vanilla, mulberry, orange, and raspberry fragrances (shorthand: V, M, O, R). How many options do you have? Solution: for the first choice you have 4 choices; for the second, 3; for the third, ; and for the last, only 1. The total ways then to select the four candles are: 4!=4 3 1 = 4. Permutation Rule: (when items are all different) We must have a total of n different items available. (This rule does not apply if some items are identical to others.) We must select r of the n items (without replacement.) Order DOES matter. ABC is NOT the same as BCA. n P r n! = ( n r)! Permutation There are eight horses in a race. In how many different ways can these horses come in first, second, and third? n = 8 r = 3 8! 8! P = = = 336 ( 8 3)! 5! 8 3 Permutations Rule ( when some items are identical to others ) If there are n items with n 1 alike, n alike,... n k alike, the number of permutations of all n items is n! n 1!. n! n k! 6

7 How many distinguishable ways can the letters in SECRETARIES be written? Number of letters: n = 11 S: E: 3 R: 11! = 1, 663, 00! 3!! Combinations Rule: We must have a total of n different items available. We must select r of the n items (without replacement.) Order DOES NOT matter. The combination ABC is the same as CBA. n C r n! = r!( n r)! Combination Rule Calculator (TI-83/84) Eight players are in a competition, top three will be selected for the next round (order does not matter). How many possible choices are there? Order does NOT matter, so with n= 8, and r = 3: 8! C = = 56 3!( 8 3)! 8 3 Math PRB : npr use this for Permutation (when all n items are different) : 10P3: enter 10 first, then npr, then 3. 3: ncr use this for Combination : 10C3: enter 10 first, then ncr, then 3. 4:! use this for factorial : 7!: enter 7 first, then the factorial sign Discrete Probability Distribution Binomial Probability Distribution Random variable has a single numerical value, determined by chance, for each outcome of a procedure. Random variables are usually denoted X or Y Chapter 5 Discrete takes on a countable number of values (i.e. there are gaps between values). Continuous there are an infinite number of values the random variable can take, and they are densely packed together (i.e. there are no gaps between values) 7

8 Would the following random variables, X, be discrete or continuous? X = the number of sales at the drive-through during the lunch rush at the local fast food restaurant. X = the time required to run a marathon. X = the number of fans in a football stadium. X = the distance a car could drive with only one gallon of gas. Random variables A probability distribution is a description of the chance a random variable has of taking on particular values. It is often displayed in a graph, table, or formula. Properties: Discrete probability distribution includes all the values of DRV; For any value x of DRV: 0 P(x) 1; The sum of probabilities of all the DRV values equals to 1; The values of DRV are mutually exclusive. An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were rated on a score from 1 to 5, where 1 was extremely passive and 5 extremely aggressive. A score of 3 indicated neither trait. The results are shown below: Frequency cont. Construct the probability distribution: Verify that it s a probability distribution. For all values of X, 0 P(x) 1 Frequency Probability P(X=x) 4/150 = /150 = 0. 4/150 = /150 = =1 1/150 = 0.14 cont. cont. Probability P(X=x) Probability P(X=x) Graph the distribution. Note: the area of each bar is equal to the probability of a particular outcome What is the probability that randomly selected worker got a score 3 or less? P(X 3) = = 0.66 What is the probability that a randomly selected worker scored at least 4? P(X 4) = = 0.34 What is the probability that a randomly selected worker did not score 5? P( X 5) = =

9 The Mean and Standard Deviation of a Discrete Random Variable The mean or expected value of a discrete random variable is given by µ = xp( x) = x1 P( x1) + x P( x ) +... The variance of a discrete random variable is given by σ = ( x µ ) P( x) = ( x1 µ ) P( x1) + ( x µ ) P( x ) +... The standard deviation is σ = σ cont. Probability P(X=x) What is the mean score? What can you conclude? µ = xp( x) = x1 P( x1) + x P( x ) µ = =. 94 We can conclude that the mean personality trait is neither extremely passive nor extremely aggressive, but is slightly closer to passive. cont. Probability P(X=x) µ =. 94 Find the variance and standard deviation of the scores. What can you conclude? σ = ( x µ ) P( x) = ( x1 µ ) P( x1) + ( x µ ) P( x ) +... σ = ( 1. 94) (. 94) 0. + ( 3. 94) ( 4. 94) 0. + ( 5. 94) 014. σ = σ = σ = Most of the data values differ from the mean by no more than 1.3 points. 9

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