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by: Sophie Levy

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# Bstats Week 4 notes MATH 1140

Marketplace > Tulane University > Math > MATH 1140 > Bstats Week 4 notes
Sophie Levy
Tulane

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r and n
COURSE
PROF.
Robert Herbert,
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Math

This 3 page Class Notes was uploaded by Sophie Levy on Friday September 23, 2016. The Class Notes belongs to MATH 1140 at Tulane University taught by Robert Herbert, in Fall 2016. Since its upload, it has received 6 views. For similar materials see Statistics For Business in Math at Tulane University.

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Date Created: 09/23/16
9/21 Counting Product Rule/Multiplication Rule: the decision can be made in n * n  * n  1  n2ways3 k  - A decision is the result of K choices  o The first choice can be made in n  ways1 o The second choice can be made in n ways, re2 rdless of the first choice  o The third choice can be made in n ways, 3 gardless of previous choices  o Etc. o The kth choice can be made in n ways, k gardless of previous choices  - Ex: we have 4 children: A, B, C, and D and we have a row of 4 seats  In how many ways may the children sit?       Seats: C – D ­  1. Select a child for seat one: n =14 (4 options) 2. Select a child for seat two: n = 3  2 3. Select a child for seat three: n =32 4. Select a child for seat four: n =11 (last option) o Answer is 4!= 4*3*2*1= 24 o 4!: 4 factorial Factorials: - n!: n factorial= n*(n­1)*(n­2)… - 0!= 1 - Ex: we have a meal with: o 4 choices of vegetable o 3 choices of salad o 2 choices of main course  o Math: 4*3*2=24 choices r Permutation: Given a pool of n objects, an r permutation is obtained by selected r distinct  elements from the pool and placing them in some specific order n - P  =r Pn= r mber of r permutations from a pool of n objects - 4 4= 4*3*2*1 - Ex: if we have 10 members in a club, in how many ways can we select a president, VP,  secretary, and treasurer  o Saying John, Jane, Jake, and Jill ≠ Jill, Jane, John, Jake  o So 1. pick president: 10 choices 2. pick VP: 9 choices 3. pick secretary: 8 choices 4. pick treasurer: 7 choices o There are 10*9*7*6= 5040 ways we can make our selections  o n rn(n­1)(n­2)…(n­r+1)                          Since you made r choices there must be r factors  - Ex: flip a fair coin 20 times and record the sequence of heads and tails  o First flip: n :12 choices o Second flip: n = 22choices 20 o Math: 2  = 10,048,000 possible outcomes   o Point: permutations do not always occur  - Ex: pool of A, B, C, D  4 o ( ) 24 choose 2” = 6 because there are 6 two element subsets from a pool of 4   all permutations include: AB, AC, AD / BC, BD / DC  (AB)=(BA) o Each 2 combination gives rise to 2 permutations  with the combination AB, we can from 2 permutations: BA or AB  R combinations: given a pool of n objects, an r­combination is obtained by selecting r distinct  objects from the pool– order does NOT matter but NO repetition  n  binomial coefficient: ( ): “n rhoose r” counts the number of r element subsets that you  can form from a set of n objects nCr= C  = ( ) = “n chose r” r r - Each r­combination gives rise to r factorial permutations  - Ex: A, B, C  o can be written down in 6 possible ways: ABC, ACB, BAC, BCA, CAB, CBA o you have 3 choices for the first letter, 2 choices for the second, 1 choice for the  lnst  o ( )*r!= nPr =  - Ex with equation: ( ) =38! / (3!)(5!) = 8*7*6*5*4*3*2*1 / (3*2*1)(5*4*3*2*1) = 56  - Ex: how many 5­card stud poker hands are possible? o ( ) =552*51*50*49*48*47*46*43… / (5!)(47*46*43…) =2,598,960  - Urn ex: there’s 4 red and 6 blue; Select 5 balls without replacement and find p(2r+3b) o The sample space ( ) = 10*5*8*7*6 / 1*2*3*4 =508  1. Pick two red balls from the pool of 4: ( ) 4 6 2 2. Pick 3 blue balls from the pool of 6: ( )  3 o Use multiplication rule: the number of 5 element samples with 2 red and 3 blue  4 6 is ( 2* ( )3= [4*3*2*1/ (2*1)(2*1)] * [6*5*4*3*2*1 / (3*2*1)(3*2*1)]   6*20=120 o p(24+3b)= 120/252=.48

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