CHAPTER 17 Ted Shi, Kevin Yen Betters, 1st PROBABILITY MODELS

WHY DO WE USE PROBABILITY? Instead of simulations, we can use actual probabilities to determine the chance of an event occurring.

WHAT IS A BERNOULLI TRIAL? Two possible outcomes on each trial: Success Failure Probability, p, is the same on each trail Each trial is independent of one another Or, if not independent, then sample must be < 10% of pop. Not usually that interesting…

WHY/HOW DO WE USE A GEOMETRIC PROBABILITY MODEL? Tell us how long, and with what probability, we can achieve success, denoted by “Geom(p)” Follows rules of Bernoulli’s trials. Formulas: P ( X = x ) = ( q x – 1 )( p ) μ = 1 / p σ = √ ( q / p2 ) p = probability of success q = 1 – p = probability of failure X = no. of trials until first success

WHY/HOW DO WE USE A BINOMIAL PROBABILITY / MODEL? Interested in number of successes (usually > 1) in a specific number of trials Not as easy as it seeeeems. Formulas: P ( X = x ) = ( n C x ) ( p x )( q n - x ) n = number of trials p = probability of success q = 1 – p = probability of failure x = no. of successes in n trials

BINOMIAL MODEL CONT. Formulas (cont.): μ = ( n )( p ) σ = √ ( n )( q )( p ) n = number of trials p = probability of success q = 1 – p = probability of failure x = no. of successes in n trials

APPLYING THE NORMAL MODEL Normal models extend indefinitely – need to consider 3 standard deviations Useful when dealing with large number of trials involving binomial models Works well if we expect at least 10 success/failures Success/Failure Condition A Binomoial model can be considered Normal if we expect np ≥ 10 and nq ≥ 10

CONTINUOUS RANDOM VARIABLES The Binomial Model is discrete (success = 1, 2, etc.) The Normal Model is continuous (any random variable at any value :o ) Continuity Correction “For continuous random variables we can no longer list all the possible outcomes and their probabilities” e.g. donating ≥ 1850 pints of blood vs donating exactly 1850 pints of blood

QUESTION 17: STILL MORE LEFTIES Original premise: Question 13. Assume that 13% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below

QUESTION 17: STILL MORE LEFTIES Our Situation: Question 17. Suppose we choose 12 people instead of the 5 chosen in Exercise 13. Conditions: Is this a Bernoulli trial? Yes. Success: LEFT, Failure: Right Independent? No, but less than 10% of population.

QUESTION 17: STILL MORE LEFTIES A) Find the mean and standard deviation of the number of right-handers in the group. P (right-handed) = 0.87 μ = ( n )( p ) σ = √ ( n )( q )( p ) = (12)(0.87) = √ (12)(0.13)(0.87) = 10.44 people = 1.164 people

QUESTION 17: STILL MORE LEFTIES B)What’s the probability that they’re not all right- handed? P (all not right-handed) = 1 – P(all right-handed) = 1 – P (X = x) = 1 – P( X = 12 right handed ppl) = 1 – (12 C 12 )(0.8712)(0.130) = 0.812 p (right handed) = 0.87, p (left handed) = 0.13

QUESTION 17: STILL MORE LEFTIES C) What’s the probability that there are no more than 10 righties? P (no more than10 righties) = P (X ≤ 10) = P (X =0) + P (X = 1) + P (X =2) ...+P (X = 10) = (120)(0.13)12 (0.87)0+(121)(0.13)11(0.87)1...+(122)(0.13)10(0.87)2 = 0.475 P (no more than10 righties) = 0.475

QUESTION 17: STILL MORE LEFTIES D) What’s the probability that there are exactly 6 of each? P (exactly 6 of each) = P(Y =6) = (126) (0.13)6 (0.87)6 = 0.00193 P (exactly 6 of each) = 0.00193

QUESTION 17: STILL MORE LEFTIES E) What’s the probability that the majority is right- handed? P (majority righties) = P(Y ≥ 7) = P (X ≤ 10) = P (X = 7) + P (X = 8) + P (X = 9) ...+P (X = 12) = (127)(0.13)5 (0.87)7 + (128)(0.13)4 (0.87)8 +... (1212)(0.13)0 (0.87)12 = 0.998 P (majority righties) = 0.998

QUESTION 19: TENNIS, ANYONE? A certain tennis player makes a successful first serve 70% of the time. Assume that each serve is independent of the others.

QUESTION 19: TENNIS, ANYONE? A) If she serves 6 times, what’s the probability she gets all 6 serves in? P (all six serves in) = P (X = 6) = (66)(0.70)6 (0.30)0 = 0.118 P (all six serves in) = 0.118

QUESTION 19: TENNIS, ANYONE? B) If she serves 6 times, what’s the probability she gets exactly 4 serves in? P (exactly four serves in) = P (X = 4) = (64)(0.70)4 (0.30)2 = 0.324 P (exactly four serves in) = 0.324

QUESTION 19: TENNIS, ANYONE? C)If she serves 6 times, what’s the probability she gets at least 4 serves in? P (at least four serves in) = P (X = 4) + P (X = 5) = P (X = 6) = (64)(0.70)4 (0.30)2 + (65)(0.70)5 (0.30)1 + (66)(0.70)6 (0.30)0 = 0.744 P (at least four serves in) = 0.744

QUESTION 19: TENNIS, ANYONE? D) If she serves 6 times, what’s the probability she gets no more than 4 serves in? P (no more than four serves in) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = (60)(0.70)0 (0.30)6 + (61)(0.70)1 (0.30)5 + (62)(0.70)2 (0.30)4 + (63)(0.70)3 (0.30)3 + (64)(0.70)4 (0.30)2 = 0.580 P (no more than four serves in) = 0.580