January 14, 2012 Lecture 3
Probability Theory Definition Mutually exclusive events: Two events A and B are mutually exclusive if A B = φ Definition Special Addition Rule: Let A and B be two mutually exclusive events, then probability of P(A or B) = P(A B) = P(A) + P(B) More generally if A, B, C,... are mutually exclusive events, then P(A or B or C or ) = P(A) + P(B) + P(C) +
Probability Theory Example Size (acres) relative frequency Event Under 10 0.084 A 10 49 0.265 B 50 99 0.161 C 100 179 0.149 D 180 259 0.077 E 260 499 0.106 F 500 999 0.076 G 1000 1999 0.047 H 2000 and over 0.035 I
Probability Theory Example The first two columns of the given table show a relative-frequency distribution for the size of farms in the United States. The U.S. Department of Agriculture compiled this information and published it in Census of Agriculture. In the third column of table, we introduce events that correspond to the size classes. For example, if a farm is selected at random, D denotes the event that the farm has between 100 and 179 acres, inclusive. The probabilities of the events in the third column of table equal the relative frequencies in the second column. For instance, the probability is 0.149 that a randomly selected farm has between 100 and 179 acres, inclusive: P(D) = 0.149. Use the given table and the special addition rule to determine the probability that a randomly selected farm has between 100 and 499 acres, inclusive.
Probability Theory Example Solution: The event that the farm selected has between 100 and 499 acres, inclusive, can be expressed as (D or E or F). Because events D, E, and F are mutually exclusive, the special addition rule gives. P(D or E or F ) =P(D) + P(E) + P(F ) =0.149 + 0.077 + 0.106 =0.332. Interpretation 33.2% of U.S. farms have between 100 and 499 acres, inclusive.
Probability Theory Definition Complimentary Events: For any event A, P(Not A) = P(A ) = 1 P(A)
Probability Theory Example We saw that the first two columns of the table from previous example provide a relative frequency distribution for the size of U.S. farms. Find the probability that a randomly selected farm has a. less than 2000 acres. b. 50 acres or more.
Probability Theory Example Solution: (a) Let J = event the farm selected has less than 2000 acres. P(J) = 1 P(not J) = 1 0.035 = 0.965. Interpretation 96.5% of U.S. farms have less than 2000 acres. (b) Let K = event the farm selected has 50 acres or more. P(not K) = P(A or B) =P(A) + P(B) =0.084 + 0.265 = 0.349. P(K) = 1 P(not K) = 1 0.349 = 0.651. Interpretation 65.1% of U.S. farms have at least 50 acres.
Probability Theory A B Definition The General Addition Rule If A and B are any two events, then P(A or B) = P(A) + P(B) - P(A and B)
Probability Theory Example Data on people who have been arrested are published by the Federal Bureau of Investigation in Crime in the United States. Records for one year show that 76.2% of the people arrested were male, 15.3% were under 18 years of age, and 10.8% were males under 18 years of age. If a person arrested that year is selected at random, what is the probability that that person is either male or under 18?
Probability Theory Example Solution: Let M = event the person obtained is male, and E = event the person obtained is under 18. P(M or E) =P(M) + P(E) P(M and E) =0.762 + 0.153 0.108 =0.807. Interpretation 80.7% of those arrested during the year in question were either male or under 18 years of age (or both).
Probability Theory Class Activity 1: Suppose that you hold 20 out of a total of 500 tickets sold for a lottery. The grand-prize winner is determined by the random selection of one of the 500 tickets. Let G be the event that you win the grand prize. Find the probability that you win the grand prize. Express your answer in probability notation. 2: A bowl contains 12 poker chips; 3 red, 4 white, and 5 blue. One of these poker chips is selected at random from the bowl. Let B denote the event that the chip selected is blue. Find the probability that a blue chip is selected, and express your answer in probability notation.
Probability Distribution Definition Random Variable A random variable is a quantitative variable whose value depends on chance or event. Discrete Random Variable A discrete random variable is a random variable whose possible values can be listed. Sum of the Probabilities of a Discrete Random Variable For any discrete random variable X, we have P(X = x) = 1.
Probability Distribution Example When a balanced dime is tossed three times, eight equally likely outcomes are possible, as shown in Table. Here, for instance, HHT means that the first two tosses are heads and the third is tails. Let X denote the total number of heads obtained in the three tosses. Then X is a discrete random variable whose possible values are 0, 1, 2, and 3. HHH HHT HTH THH HTT THT TTH TTT
Probability Distribution Example (a) Use random-variable notation to represent the event that exactly two heads are tossed. Solution: {X=2} (b) Determine P(X = 2). Solution: P(X = 2) = 3 8 = 0.375 Interpretation There is a 37.5% chance of obtaining exactly two heads in three tosses of a balanced dime.
Probability Distribution Example (c) Find the probability distribution of X. Solution: No. of heads x Probability P(X = x) 0 0.125 1 0.375 2 0.375 3 0.125 1.000
Probability Distribution Example (d) Use random variable notation to represent the event that at most two heads are tossed. Solution: {X 2} (e) Find P(X 2). Solution: {X 2} = {X = 0} or {X = 1} or {X = 2} P(X 2) =P(X = 0) + P(X = 1) + P(X = 2) =0.125 + 0.375 + 0.375 =0.875. Interpretation There is an 87.5% chance of obtaining two or fewer heads in three tosses of a balanced dime.
Probability Distribution Class Activity The Television Bureau of Advertising, Inc., publishes information on color television ownership in Trends in Television. Following is a probability distribution for the number of color TVs, Y, owned by a randomly selected household with annual income between $15,000 and $29,999. y 0 1 2 3 4 5 P(Y=y) 0.009 0.376 0.371 0.167 0.061 0.016 Use random variable notation to represent each of the following events. The households owns
Probability Distribution Class Activity (a) at least one color TV. (b) exactly two color TVs. (c) between one and three, inclusive, color TVs. (d) an odd number of color TVs.
Probability Distribution Class Activity Use the special addition rule and the probability distribution to determine (e) P(Y 1) (f) P(Y = 2) (g) P(1 Y 3) (h) P(Y = 1, 3, 5)
Probability Distribution Class Activity A certain couple is equally likely to have either a boy or a girl. If the family has four children, let X denote the number of girls. (a) Identify the possible values of the random variable X. (b) Determine the probability distribution of X. (Hint: There are 16 possible equally likely outcomes. One is GBBB, meaning the first born is a girl and the next three born are boys.)
Probability Distribution Class Activity Use random-variable notation to represent each of the following events. Also use the special addition rule and the probability distribution obtained in part (b) to determine each event s probability. The couple has (c) exactly two girls. (d) at least two girls. (e) at most two girls. (f) between one and three girls, inclusive. (g) children all of the same gender.
DECISION MAKING WITH PROBABILITIES Definition Mean of a Discrete Random Variable The mean of a discrete random variable X is denoted by µ or E(X ) and defined by E(X ) = µ = x P(X = x) The terms expected value and expectation are commonly used in place of the term mean of random variable.
DECISION MAKING WITH PROBABILITIES Example Prescott National Bank has six tellers available to serve customers. The number of tellers busy with customers at, say, 1:00 P.M. varies from day to day and depends on chance; hence it is a random variable, say, X. Past records indicate that the probability distribution of X is as shown in the first two columns of table given. Find the mean of the random variable X.
DECISION MAKING WITH PROBABILITIES Example x P(X=x) x P(X = x) 0 0.029 0.000 1 0.049 0.049 2 0.078 0.156 3 0.155 0.465 4 0.212 0.848 5 0.262 1.310 6 0.215 1.290 4.118 E(X ) = µ = x P(X = x) = 4.118 Interpretation The mean number of tellers busy with customers is 4.118.
DECISION MAKING WITH PROBABILITIES Key Fact Interpretation of the Mean of a Random Variable In a large number of independent observations of a random variable X, the average value of those observations will approximately equal the mean, µ, of X. The larger the number of observations, the closer the average tends to be µ
DECISION MAKING WITH PROBABILITIES Definition Standard Deviation of a Discrete Random Variable The standard deviation σ of the random variable can be obtain from the following formula: σ = {x 2 (PX = x)} µ 2
DECISION MAKING WITH PROBABILITIES Example Use the data of previous example and determine the standard deviation of the discrete random variable X. x P(X=x) x 2 x 2 P(X = x) 0 0.029 0 0.000 1 0.049 1 0.049 2 0.078 4 0.312 3 0.155 9 1.395 4 0.212 16 3.392 5 0.262 25 6.550 6 0.215 36 7.740 19.438
DECISION MAKING WITH PROBABILITIES Example σ = {x 2 (PX = x)} µ 2 = 19.438 4.118 2 = 1.6 Interpretation Roughly speaking, on average, the number of busy tellers is 1.6 from the mean of 4.118 busy tellers.
DECISION MAKING WITH PROBABILITIES Class Activity A factory manager collected data on the number of equipment breakdowns per day. From those data, she derived the probability distribution shown in the following table, where W denotes the number of breakdowns on a given day. w 0 1 2 P(W = w) 0.80 0.15 0.05 (a) Determine µ and σ (b) On average, how many breakdowns occur per day? (c) About how many breakdowns are expected during a 1-year period, assuming 250 work days per year?
DECISION MAKING WITH PROBABILITIES Example You may have two games that lead to different random variables. These two random variables are listed here. Game 1 x in $ P(X = x) 100 0.5 0 0.5 Game 2 x in $ P(X = x) 5 75 6 1-25 6 Situation 1: The random variable for game 1 can describe a game in which a coin is tossed. You win $100 if heads comes up, otherwise you win nothing. Situation 2: This random variable can also describe a business situation. If the item is sold, you get $100 profit. There is a 50% chance that the item will be sold. If the item is not sold, it may be returned to the supplier, at no cost.
DECISION MAKING WITH PROBABILITIES Key Fact Expected Value Criterion for Making a Decision In a game If E(x) > 0 play the game, you should win in the long run. If E(x) < 0 do not play. If E(x) = 0 we call this a fair game.
DECISION MAKING WITH PROBABILITIES Example Solution: Game 1: Game 2: E(X ) = 100 0.5 + 0 0.5 = 50 E(X ) = 75 5 6 25 1 6 = 58.33 Using the expected value criteria, choose game 2.
DECISION MAKING WITH PROBABILITIES Class Activity Game 1 x P(X = x) -1 3 8 0 1 8 2 3 8 7 1 8 Which game do you choose? Game 2 x P(X = x) 0 2 52 1 2 52 2 40 52 3 8 52 (a) If you want to maximize the chance of winning something (b) If you want to minimize the chance of losing (c) If you want the highest expected value