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# Permutation, Combination, Binomial Distribution CH 15 WEEK 6 NOTES STAT 2332

Marketplace > University of Texas at Dallas > Statistics > STAT 2332 > Permutation Combination Binomial Distribution CH 15 WEEK 6 NOTES
veronica
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Notes over Ch. 15 Week 6 (factorials, permutations, combinations, and binomial probability distributions). Includes examples and formulas and definitions. 4 simple but informative pages. This mater...
COURSE
Introductory Statistics for Life Sciences
PROF.
TYPE
Class Notes
PAGES
4
WORDS
CONCEPTS
Statistics, Math, Stats, intro to statistics, stat2332, Probability, Permutations, Combinations, binomial, factorial, factorials, notes, #statnotes
KARMA
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This 4 page Class Notes was uploaded by veronica on Saturday October 8, 2016. The Class Notes belongs to STAT 2332 at University of Texas at Dallas taught by Dr. Chattopadhyay in Spring 2016. Since its upload, it has received 13 views. For similar materials see Introductory Statistics for Life Sciences in Statistics at University of Texas at Dallas.

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Date Created: 10/08/16
NOTES Week 6 Ch. 15: Permutations. Combinations. The Binomial Distribution factorial: 0! = 1 1! = 1 2! = 2×1= 1 3! = 3×2×1= 6 and so on Permutation: ordered arrangement of “n” distinct items; how many ways we can rearrange a set in order. Ex: {a,b,c} can be rearranged as{b,a,c}{b,c,a}{a,c,b}{c,a,b}{c,b,a}  You can find how many ways to rearrange elements by using factorials. Ex: Since the above example had 3 items (a,b,c) we can calculate how many ways to rearrange those items by using 3! 3! =3×2×1= 6 ways to rearrange (a,b,c)  r-permutation: an ordered arrangement of r-items taken from the set of n-items; choice of r things from a set of n things where order matters. Notation/formula for # of r-permutations: n = total # of items in sample P(n,r) or n r r = # of items we’re looking for Example Problem: From a club of 18 members, a President, Vice-President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? Solution: n=18, r=5 (# of offices (VP, secretary, etc.) to be filled) 18! 18! 18 5 = (18−5)!= 13!= 18 × 17 × 16 × 15 × 14 = 1,028,160 Combination: an arrangement of things selected in which order of selection doesn’t matter. Notation/formula to find the number of combinations of n items chosen r at a time: Permutations are for lists (order matters) C(n,r) orn r and combinations are for groups (order doesn’t matter). Combination Ex: Choose 3 desserts from a menu of 10 C(10,3) = 120 **Note: We will be using k instead of r in class; replace r with k for combinations. Example Problem: A student must answer 4 out of 5 essay questions on a test. How many different ways can the student select the questions? Solution: n=5, r=4 5! 5! 5×4! 5 4 = 4!(5−4)!= 4!1!= 4! = 5 Binomial probability distribution We’re basically finding the probability of “success” aka what we’re looking for Binomial coefficient: (see formula below; this formula is the same as combination formula) Note: (n-k)! ≠ n! – k! Binomial Assumptions/Rules: “success”  p or 1. Each observation falls into 1 of 2 categories “failure”  1-p 2. Fixed n 3. Observations are independent 4. Probability of “success” p stays constant for each trial Specifically, k = # of successes n = # of observations AKA total # of trials p = probability of success for each trial 1−p = probability of failure for each trial The probability of k successes out of n trials is ????! ???? ????−???? ???? ???? = ????! ???? − ???? !) ???? ???? − ???? ) Ex: Multiple Choice Quiz (4 choices each). Student guesses 5 questions n=5 Success=getting it right p=0.25 (a) Find the probability he gets 3 right ????! ???? ???? = ???? =) (????.????????) ????.???????? )????−???? = ????.???????????????? ????! ???? − ???? !) (b)Find the probability he gets at least three right ???? ???? ≥ ???? = ???? ???? = ???? + ???? ???? = ???? + ????(???? = ????) (c) Find the probability he gets more than three right ???? ???? > ???? = ???? ???? = ???? + ????(???? = ????) Example binomial problem: Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover? Solution: Successthe patient recovers Failurethe patient dies k = number who recover = 4 n = 6 p = 0.25 (1-p) = 0.75 (a) Probability that 4 will recover = ????! ???? ???? ???? ???? = ???? = ????! ???? ! (????.????????) ????.???????? ) = ????.???????????????? (b)Probability that less than 4 will recover = ???? ???? < ???? = ???? ???? = ???? + ???? ???? = ???? + ???? ???? = ???? + ????(???? = ????) (c) Probability that at least 1 will recover = ???? ???? ≥ ???? = ???? ???? = ???? + ???? ???? = ???? + ⋯+ ????(???? = ????) or use the “at least one” rule 1-P(none) ???? − ????(???? = ????) Thank you for reading! Next part will cover geometric probability distribution, poisons, and exponential distributions. A study guide for Exam 2 will also be uploaded.

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