Chapter 6: Probability and Simulation The study of randomness
Introduction Probability is the study of chance. 6.1 focuses on simulation since actual observations are often not feasible. When we produce data by random sampling or randomized comparative experiments laws of probability answer the question what would happen if we did this many times? Probability is the basis of inference
6.1 Simulation The experiment consists of spinning the spinner three times and recording the numbers as they occur (e.g. 123). We want to determine the proportion of times that at least one digit occurs in its correct position. (we will use a calculator instead). Guess the proportion of times at least one digit will occur in its proper place.
Get your calculators out!! Enter the command randint(1,3,3) Continue to press enter to generate more three-digit numbers Record the result in a table At least one digit in the correct position None of the digits in the correct position
The more repetitions, the closer a result s occurrence will get to it s true likelihood. Independence: When the result of one trial (coin toss, dice roll) has no effect or influence on the next toss.
Coin Toss
Simulation Steps 1. State the problem or describe the random phenomenon Ex: Toss a coin 10 times, what is the likelihood of a run of at least 3 consecutive heads or 3 consecutive tails? 2. State the Assumptions (there are 2) A head or a tail is equally likely to occur on each toss Tosses are independent of each other 3. Assign digits to represent outcomes (want efficiency) In a random # table, even and odd digits occur with the same long-term relative frequency (50%) One digit simulates one toss of the coin Odd digits represent heads; even digits represent tails Successive digits in the table simulate independent tosses 4. Simulate many repetitions Looking at 10 consecutive digits in Table B simulates one repetition. Read many groups of 1- digits from the table to simulate many rep
Be sure to keep track of whether or not the event we want (a run of at least 3 heads or at least 3 tails) occurs on each repetitions Here are the first 3 repetitions starting at line 101 in Table B. Digits: 19223 95034 05756 28713 96409 12531 H/T: HHTTH HHTHT THHHT TTHHH HTTTH HTHHH Run of 3: YES YES YES 22 repetitions were done for a total of 25. 23 of them did have a run of 3 or more Heads or tails. 5. State your conclusions We estimate the probability of a run of size 3 by the proportion 23/25 =.92 *Of course 25 reps are not enough to be confident that our estimate is accurate so we can tell a computer to do thousands of repetitions (or TRIALS) for us. A long simulation finds that the true probability is.86
Assigning digits Some ways more efficient than others. Example: Choose a person at random from a group of which 70% are employed. One digit simulates one person: 0, 1, 2, 3, 4, 5, 6 = employed 7, 8, 9 = not employed 00-69 employed and 70-99 not employed could also have worked, but is less efficient b/c requires twice as many digits and ten times as many numbers. Example 2: Choose one person at random from a group of which 73% employed Now 00-72 = employed, 73-99 = not employed Example 3: Choose one person at random from a group of which 50% are employed, 20% are unemployed, and 30% are not in the labor force: 0-4 = employed, 5-6 = unemployed, 7-9 = not in labor force.
Frozen Yogurt Sales example Orders of frozen yogurt flavors (based on sales) have the following relative frequencies: 38% chocolate, 42% vanilla, 20% strawberry. We want to simulate customers entering the store and ordering yogurt. How would you simulate 1- frozen yogurt sales based on recent history using table?
Randomizing with Calculator Block of 5 random digits from table Rolling a die 7 times 10 numbers from 00-99
0 1 0.6 0.01
You read in a book on poker that the probability of being dealt three of a kind in a five-card poker hand is 1/50. Explain in simple language what this means. If the hands were dealt many times, about 2% (1out of 50) hands will contain a three of a kind.
6.2 Probability Models Proportion of heads to tails in a few tosses will be erratic but after thousands of tosses will approach the expected.5 probability
Probability models have two parts: A list of possible outcomes A probability for each outcome.
Sample Space To specify S we must state what constitutes an individual outcome, then which outcomes can occur (can be simple or complex) Ex: coin tossing, S = {H, T} Ex: US Census: If we draw a random sample of 50,000 US households, as the survey does, the S contains all 50,000
Rolling two dice At a casino- 36 possible outcomes when we roll 2 dice and record the up-faces in order (first die, second die) Gamblers care only about number of dots face up so the sample space for that is: S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Techniques for finding outcomes 1. Tree diagram For tossing a coin then rolling a die
2. Multiplication Principle 2x6 = 12 for same example 3. Organized list: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6
With/without replacement If you take a card from a deck of 52, don t put it back, then draw your 2 nd card etc., that s without replacement. Ex: how many different 3 digit numbers can you make: 10x9x8 = 720 If you take a card, write it down, put it back, draw 2 nd card etc., that s with replacement. Ex: 10x10x10 = 1000
Probability Rules 1. Any probability is a number between 0 and 1 2. The sum of the probabilities of all possible outcomes = 1 3. If 2 events have no outcomes in common (they can t occur together), the probability that one OR the other occurs is the sum of their individual probabilities Ex: If one event occurs in 40% of all trials, another event happens in 25% of all trials, the 2 can never occur together, then one or the other occurs on 65% of all trials 4. The probability that an event doesn t occur is 1 minus the probability that it does occur Ex: If an event happens in 70% of all trials, it fails to occur in the other 30%
Venn diagrams help! Ex: Probability of rolling a 5? B/c P(roll a 5 with 2 die) = P(1,4) + P(3,2) + P(2,3) + P(4,1) = 1/36 + 1/36 + 1/36 + 1/36 = 1/9 or.111
Independence & the Multiplication Rule To find the probability for BOTH events A and B occurring Example: Suppose you plan to toss a coin twice, and want to find the probability of rolling a head on both tosses. A = first toss is a head, B = second toss is a head. So (1/2)(1/2) = ¼. We expect to flip 2 heads on 25% of all trials. The more times we repeat this, the closer our average probability will get to 25%. The multiplication rule applies only to independent events; can t use it if events are not independent!
Independent or not? Coin toss I: Coin has no memory and coin tossers cannot influence fall of coin Drawing from deck of cards NI: First pick, probability of red is 26/52 or.5. Once we see the first card is red, the probability of a red card in the 2 nd pick is now 25/51 =.49 Taking an IQ test twice in succession NI
More applications of Probability Rules If two events A and B are independent, then their complements are also independent. Ex: 75% of voters in a district are Republicans. If an interviewer chooses 2 voters at random, the probability that the first is a Republican and the 2 nd is not a republican is.75 x.25 =.1875
6.3 General Probability Rules
Addition Rule for Disjoint events
General Addition rule for Unions of 2 events
Example: Deb and Matt are waiting anxiously to hear if they ve been promoted. Deb guesses her probability of getting promoted is.7 and Matt s is.5, and both of them being promoted is.3. The probability that at least one is promoted =.7 +.5 -.3 which is.9. The probability neither is promoted is.1. The simultaneous occurrence of 2 events (called a joint event, such as deb and matt getting promoted) is called a joint probability.
Conditional Probability The probability that we assign to an event can change if we know some other event has occurred. P(A B): Probability that event A will happen under the condition that event B has occurred. Ex: Probability of drawing an ace is 4/52 or 1/13. If your are dealt 4 cards and one of them is an ace, probability of getting an ace on the 5 th card dealt is 3/48 or 1/16 (conditional probability- getting an Ace given that one was dealt in the first 4).
In words, this says that for both of 2 events to occur, first one must occur, and then, given that the first event has occurred, the second must occur.
Remember: B is the event whose probability we are computing and A represents the info we are given.
Extended Multiplication rules The union of a collection of events is the event that ANY of them occur The Intersection of any collection of events is the event that ALL of them occur
Example Only 5% of male high school basketball, baseball, and football players go on to play at the college level. Of these only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years. Define these events: A = competes in college B = competes pro C = pro career longer than 3 years P(A) =.05 P(B A) =.017 P(C A and B) =.400 What is the probability a HS athlete will have a pro career more than 3 years? The probability we want is therefore P(A and B and C) = P(A)P(B A)P(C A and B) =.05 x.017 x.40 =.00034 So, only 3 of every 10,000 high school athletes can expect to compete in college and have a pro career of more than 3 years.
Extended tree diagram + chat room example 47% of 18 to 29 age chat online, 21% of 30 to 49 and 7% of 50+ Also, need to know that 29% of all internet users are 18-29 (event A1), 47% are 30 to 49 (A2) and the remaining 24% are 50 and over (A3). What is the probability that a randomly chosen user of the internet participates in chat rooms (event C)? Tree diagram- probability written on each segment is the conditional probability of an internet user following that segment, given that he or she has reached the node from which it branches. (final outcome is adding all the chatting probabilities which =.2518)
Bayes Rule Another question we might ask- what percent of adult chat room participants are age 18 to 29? P(A1 C) = P(A1 and C) / P(C) =.1363/.2518 =.5413 *since 29% of internet users are 18-29, knowing that someone chats increases the probability that they are young! Formula sans tree diagram: P(C) = P(A1)P(C A1) + P(A2)P(C A2) + P(A3)P(C A3)
6.3 Need to Know summary(print) Complement of an event A contains all outcomes not in A Union (A U B) of events A and B = all outcomes in A, in B, or in both A and B Intersection(A^B) contains all outcomes that are in both A and B, but not in A alone or B alone. General Addition Rule: P(AUB) = P(A) + P(B) P(A^B) Multiplication Rule: P(A^B) = P(A)P(B A) Conditional Probability P(B A) of an event B, given that event A has occurred: P(B A) = P(A^B)/P(A) when P(A) > 0 If A and B are disjoint (mutually exclusive) then P(A^B) = 0 and P(AUB) = P(A) + P(B) A and B are independent when P(B A) = P(B) Venn diagram or tree diagrams useful for organization.