tues. feb 1

Morgan Carter

1

February 1, 2011

ST371-‐02

Professor Smith

Class Summary Review from last session:

€

nn1,n2...nk

⎛

⎝ ⎜

⎞

⎠ ⎟ =

n!n1!n2!...nk! where

€

n = n1 + n2 + ...+ nk

This can be used for the # of ways to rearrange the letters in a word where there are

n letters and k types of letters (A, B, C, etc)

A general formula for combinations like this is:

€

Ck,n =nk⎛

⎝ ⎜ ⎞

⎠ ⎟ =

n!k!(n − k)!

Think of a deck of cards so k is the number chosen out of n cards making (n-‐k) the

number not chosen.

Reviewed the 9 people in 3 cars example from past worksheet to further examine.

Scrabble Problem:

Given the scrabble tiles:

a. How many ways can the tiles be arranged with no constraints?

b. How many ways can the tiles be arranged so that the word begins with a

consonant?

c. How many ways can the tiles be arranged so that the word ends with a

vowel?

d. How many ways can the tiles be arranged so that the consonants and vowels

alternate?

e. Repeat question d but add in this tile:

W H I T E

O

Morgan Carter

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f. How many ways can the tiles be arranged if all the vowels must be to the left

of the consonants?

Solutions:

a. simply 5! or in the context of the review from last class 5!/(1!1!1!1!1!)

because each letter is unique

b. (Visually) 3 4 3 2 1 are the possible choices for each position given that

there are three consonants contained within the sample. 3*4*3*2*1=72

c. (Visually) 4 3 2 1 2 are the possible choices for each position given that

there are two vowels contained within the sample. Note that this is similar to

b except that the final position is the constricting one. 4*3*2*1*2=48

d. (Visually) 3 2 2 1 1 are the possible choices for each position given that

there are three consonants and two vowels (so it must start and end with a

consonant). 3*2*2*1*1=12

e. (Visually) 6 3 2 2 1 1 are the possible choices for each position. The first

position can be a vowel or consonant but decides of which group the other

positions will be. 6*3*2*2*1*1=72

f. (Visually) 3 2 1 3 2 1 are the possible choices for each position. The left

side contains the vowels (1-‐3) and the right side contains the consonants (1-‐

3). 3!*3!=36

Worksheet handed out and discussed – see page 4

2nd worksheet passed out and discussed – see page 5

Morgan Carter

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NEW TOPIC SECTION: Conditional Probability and Independence

Readings: 2.4 and 2.5 in Devore and Ch. 3 in Cartoon Guide

Pr(B|A) = the probability of B given that A has happened Example: probability that a card is a heart (B) given that it is red (A)

€

Pr(B | A) =Pr(B∩ A)Pr(A)

≠ Pr(B) Sometimes it does equal Pr(B) but it does not have to If you add in a C circle which overlaps A and B, conditions such as the following can occur:

€

Pr(C | (A∩ B)) =Pr(A∩ B∩C)Pr(A)Pr(B | A)

You muse include every branch passed. Note: You cannot condition on a null set

A B A

B

A’ B

B’

B’

Morgan Carter

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Morgan Carter

5

Amber Purvis ST371-002

Class Summary for Tuesday, February 1, 2011

• First, we reviewed the car problem

9!(𝐶1)! (9 − 𝐶1)!

∗ �(9 − 𝐶1)!

𝐶2! (9 − 𝐶1 − 𝐶2)!� ∗

(9 − 𝐶1 − 𝐶2)!𝐶3! 0!

=9!

𝐶1!𝐶2!𝐶3!

from last Thursday.

•

Starting with the letters: WHITE

Scrabble Game Problem

a) How many ways of arranging the letters? 5! Ways

b) Event b = begins with a vowel (a,e,i,o,u)

2 * 4 * 3 * 2 *

c) Event c = begins with a consonant.

1

3 * 4 * 3 * 2 * 1

d) Event d = ends in a vowel.

4 * 3 * 2 * 1 * 2 Start with the last place.

e) Event e = Vowel and consonants alternate.

3 * 2 * 2 * 1 * 1

f) Add an O to the mix: WHITEO

3 3 2 2 1 1

3 3 2 2 1 1 2(3!)(3!)

g) All vowels to the left of consonants.

3 2 1 3 2 1

• Went over “Solutions – Problem Set 2” Handout.

• Went over “Poker” Handout.


• Started Section 2.4-5:

Conditional Probability and Independence

B -A and B (1) A B’ -A and B’ (2) B -A’ and B (3) A’ B’ -A’ and B’ (4)

B B’ A A and B (1) A and B’ (2) A’ A’ and B (3) A’ and B’ (4)

Conditional Probability:

Pr(B|A) = Probability of B given that A already happened.

Pr(𝐵|𝐴) =Pr(𝐵 𝑎𝑛𝑑 𝐴)

Pr(𝐴)

2 3 1

4

A B

Chelsea Ratzlaff

ST 371, Dr. Smith- Lecture 2/1/2010

HOMEWORK ZERO-SUMMARY OF CLASS 2/1/2011

During the first part of class we reviewed the basic method for finding the probabilities of

certain combinations. For example, the number of subsets of size k out of n objects was defined as

,!

!( )!k n

n nCk k n k

= = −

To illustrate this formula we looked at the car filling example from the previous class. Here there were 9

people trying to fir in 3 cars, cars denoted by C1, C2, C3 with all cars going and at least 1 person is in

each car. To find these possibilities we used the formula:

9 9 # 1 9 (# 1 # 2 )# 1 # 2 # 3

9! (9 # 1)!# 3!(# 1)!(9 # 1)! (# 2)!(9 # 1 # 2)!

C C CC C C

CC C C C C C

− − − =

−

•− − −

After this, we also looked at an example using scrabble. We found how many ways we can arrange the

word white. We used the representation:

_ _ _ _ _1 2 3 4 5s s s s s

Which is easily seen as represented at 5! for the 5 letters. We also found how many ways it could begin

with a vowel. We then used 2!4!3!2!1! to calculate this, considering there were only 2 possible vowels

for it to begin with. Then we found how many ways it can end with a vowel, which was the same but the

2! was at the end. Next we found how many ways we could make the vowels and consonants alternate.

The placement of the factorials look like this: 3! 2! 2! 1! 1!. We then figured the same question out if we

added an O. The answer for this was 3! 3! 2! 2! 1! 1! +3! 3! 2! 2! 1! 1!.

After this we looked at some sample problems from worksheet “Solutions Problem set 2”. We

followed the worksheet throughout which were similar to examples above. Then we received the POKER

worksheet. This worksheet explained the chances of getting certain hands in poker and how to calculate

them.

Next we started a new topic called Conditional Probability and Dependence which is sections 2-4

and 2-5 in Devore. We learned the new notation:

Pr( )Pr( / ) which is not necessarily =Pr(B)Pr( )B AB A

A∩

≡

We drew some probability trees to illustrate this concept. We also identified that a constraint on these

problems is the necessity that A must be possible for this concept to hold. That concluded class for this

day.

Byron Beddingfield ST 371 HW 0: Lecture Summary

This week in class we discussed conditional probability and independence. Conditional probability is the probability of an event given that some event already happened. The notation for this is P(A|B). This translates to the probability of A given that B has already happened. We analyzed a Venn-Diagram and a probability tree to highlight how this worked. Mathematically conditional probability is: P(A|B) = P(A&B) / P(B). We worked on an example where we had three unfair coins and were given a few of the probabilities. We then used our understanding of conditional probability to fill in the tree.

Next, we looked at independence. For independence, we used conditional probability. Independence is defined as P(A|B) = P(A). That is, the probability of A is not affected by (is independent) of B happening. We looked at a stack of cards. An example of a dependant relationship are P(card is diamond | card is red). An independent relationship is P(card is jack | card is red). Another key idea was that if A and B are mutually exclusive and they are not NULL then A and B are independent. Again we analyzed a tree and labeled the parts to highlight the concept of independence.

This lecture explored conditional probability and independence. We analyzed many examples and cases that went over several situations and how they fit into this concept of statistical analysis.

Lecture summary At the beginning of the lecture, we started with reviewing the equations of combination by continuing the previous example. Ck,n=n!/(k!(n-k)!) The next example is about scrabble with the word WHITE. A) the total number of ways to combine the five letters, the answer would be 5! B) event A: the word begins with vowel, there are a total of two vowels in this word, the other letters do not have any restrictions. Therefore, the answer would be 2*4*3*2*1 c) event C= the word ends with a vowel. The last letter has two choices, the rest of the letters are not restricted, the five digits would be 4 3 2 1 d) vowels and consonants alternate. The word must begin with a consonant, if it begins with a vowel, there will not be enough vowels to make up the word. Therefore, the five letters are 3 2 2 1

2

d’) add an “o” to the word, then let the vowels and consonants alternate. The word can begin with either with a vowel or consonant. The answer should be 3*3*2*2*1*1+3*3*2*2*1*1 for beginning with vowel or consonant.

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e) all vowels left to the consonants. There are 3 vowels and 3 consonants 3*2*1*3*2*1 Then we talked about conditional probability and independence. The tree table states the relationship

this equation can calculate the the probability of E given that F has already happened. Short example about probability B=card is red A1=card is diamond Pr(A1)=1/4 P(A1|B)=(1/4)/(1/2)=1/2 A2=card is jack P(A2|B)=(2/52)/(26/52)=1/13 Then we did an example of filling out a sequential tree table for the experiment of independently tossing three different coins (not necessary fair). The notation: H1 means a head on the first coin, T3 means a tail on the third coin a) find P(H3|H1ANDT2) b)find P(T2|H1) c) the probability of the outcome T1H2H3

Bishal Sharma Feb 1, 2011

𝑛!𝑛1!𝑛2!𝑛2!𝑛𝑛!

This is read as # of subsets of size k out of n objects

This is a more sequential way of thinking.

Let’s do an example with scrabble:

WHITE

a) How many ways can we arrange the letters? 5!

b) Event A= Begins with a Vowel: 2 4 3 2 1

( 2 is first because there is only 2 vowels to start it off)

c) Event B= begins with consonant: 2 4 3 2 1 3 4 3 2 1

d) Ends in a Vowel: 4 3 2 1 2

e) Vowel & consonants alternate: 3 2 2 2 1

f) Add letter 0 --- 3 3 2 2 1 1 + 3 3 2 2 2 1

because we have 2 different possibilities with the same outcome.

g) All vowels to left of all consonants: 3 2 1 3 2

1

Bishal Sharma Feb 1, 2011

Sometimes it is easier to find the complement of something rather the the question itself.

Lecture Summary

Tuesday, February 1st

Daniel C. Pugh

, 2011

We started class with some background equations, including the following:

�𝑛

𝑛1,𝑛2,𝑛3,𝑛4� ≡

𝑛!𝑛1!𝑛2!𝑛3!𝑛4!

Where 𝑛 = 𝑛1 + 𝑛2 + 𝑛3 + 𝑛4.

Also,

𝐶𝑘,𝑛 = �𝑛𝑘� ≡ 𝑛!

𝑘! (𝑛 − 𝑘)!

For examples, we considered the possible combinations of letters that could be made with a certain set of letters, similar to the game Scrabble. (a) We first considered the word ‘WHITE’. So with five letters, there would be five spaces for letters. As shown below, there are five possible letters to fill the first spot, four possible letters left to go in the second spot, three for the third, two for the fourth, and only one remaining possibility for the fifth spot. 5 4 3 2 1 Therefore there would be 5x4x3x2x1 = (5!) = 120 possible combinations for the five letters in ‘WHITE’. (b) Next we considered the number of ways ‘WHITE’ could be arranged so that begins with a vowel. Since there are two vowels in the word, there are two possible letters for the first spot, and then four possible for the next spot, then three, two, and one. Only the first spot changes. 2 4 3 2 1 Similarly, if the combinations were to only begin with consonants, there would be three different possibilities for the first letter. 3 4 3 2 1 (c) If you wanted the result to end in a vowel, you would set aside two possibilities for the last spot, and then four for the first spot, three for the second, two for the third, and then one for the fourth spot: 4 3 2 1 2 (d) If you wanted the word to alternate between consonants and vowels (keep in mind that with 3 consonants and 2 vowels, the first letter must be a consonant), you would have three possibilities for the first spot, a choice of two vowels for the second spot, two possible consonants for the third spot, then one vowel possibility for the fourth spot, and one remaining consonant possibility for the last spot: 3 2 2 1 1

(d’) For this example, we repeated the previous example, except this time we added another vowel to make 3 consonants and 3 vowels. So say or choice of letters was now ‘WHITEO’, and our requirements are to place the letters with alternating consonants and vowels. If the first letter was to be a consonant, you would have the choice of 3 consonants, then the choice between 3 vowels, then 2 consonants to choose from, then 2 vowels, then 1 consonant, and then 1 last vowel. You could also start the set with a vowel, which would have the same number of possible combinations as the other set. Adding these two sets together gets your total number of possible consonant/vowel alternating sets. 3 3 2 2 1 1 + 3 3 2 2 1 1 = 3x3x2x2x1x1 + 3x3x2x2x1x1 = 72 Alternatively, you could imagine there being 6 possibilities for the first spot, being either vowel or consonant, and whether or not it’s vowel or consonant, there will then be 3 possibilities for the second spot, 2 for the third, 2 for the fourth, 1 for the fifth and 1 for the sixth spot. 6 3 2 2 1 1 = 6x3x2x2x1x1 = 72 (e) If you wanted all of the vowels to be to the left of all of the consonants, you would have 3 choices of vowels for the first spot, then 2 vowels to choose from for the second spot, and then one remaining vowel for the third spot. Then all of the consonants would follow, so you would have 3 consonant choices for the fourth spot, 2 for the fifth, and 1 for the last spot. 3 2 1 3 2 1 = (3!)(3!) = 36 The possibility of this happening randomly is the number of possible combinations (36) divided by the total number of combinations of the 6 letters, which is (6!). 36/(6x5x4x3x2x1) = 0.05 There is a 5% chance of this happening randomly.

Our next lesson to learn is that the probability of a combination of things happening is equal to the sum of each event happening. For example, the probability of ( A n B’ n C ) is equal to the probability of A; times the probability of B’ given A; times the probability of C given A and B’. So following this, Pr(A n B’ n C) = ½ x ½ x ½ = 1/8.

Next we went over a printout of probability examples. The first example considered a test or quiz in which there were 5 questions, each with 4 multiple choice answers, with only one right answer each. (a) To calculate the total number of different ways the test could be answered, we multiply the number of ways to answer each question: n1 x n2 x n3 x n4 x n5 = 4 x 4 x 4 x 4 x 4 = 45 = 1024 (b) To calculate the number of ways the test could be answered without getting a single answer correct, you multiply the number of wrong answers for each question. So for this case: n1 x n2 x n3 x n4 x n5 = 3 x 3 x 3 x 3 x 3 = 35 = 243 So if you were to guess on every question of the test the chances of getting every question wrong is 243/1024, or 23.73%.

Example: 6 people are in a line to get on a bus. (a) These 6 people are like 6 distinguishable objects, so the number of ways the line could be arranged is: 6 5 4 3 2 1 => 6x5x4x3x2x1 = 6! = 720 (b) Next we considered what if 3 of the people insisted on following each other onto the bus. To figure out how many ways this could happen, we look at the number of configurations there are with 3 people together: x x x _ _ _ , _ x x x _ _ , _ _ x x x _ , _ _ _ x x x so 4 configurations. # ways = (# configurations) (orderings of 3 people together) (order of other 3 people) = 4 x 3! x 3! = 144 alternatively, you could imagine the 3 people together as one unit or “superslot” and arrange the 4 objects (3 regular slots and 1 superslot) and then arrange the three within the superslot. # ways = (arrange 4 objects) (arrange 3 people in superslot) = 4! 3! = 144 (c) For this one, there were 2 people who did not want to be together in line for the bus. So for this, the only configurations that would not be acceptable would only be the configurations in which the 2 people were together. Therefore, if you calculate the number of ways the 2 people can be together, and then subtract that from the total number of possible 6-person configurations (720, from part (a)), you will get the number of possibilities for this scenario. This is the number of ways the 2 people could be together (using the superslot method): (arrange 5 objects: 4 people and superslot) (arrange 2 people in superslot) = 5! 2! = 240 So the number of possibilities in (c) = 6! – (5! 2!) = 720 – 240 = 480 You could also use the same method used in the first part of (b), finding the number of configurations in which two slots would not be next to each other. This comes out to be 10 configurations, and then multiply that by the number of ways to arrange the 2 people and then the 4 other people. This comes out to be 10 x 2! x 4! = 480

Example

Next we looked at another handout and calculated the probability of getting dealt various poker hands. (a) First we considered a full house, which consists of a set of 3 cards of one denomination and a set of 2 cards of the same denomination (Ex. 5 of hearts, 5 of clubs, 5 of diamonds, 9 of

hearts, 9 of spades). The number for the three-of-a-kind can be chosen �131 � different ways.

Then after the number has been decided, the requisite suits can be chosen �43� ways. Applying

the same reasoning to the pair, the number can be chosen �121 � ways, and the suits can be

chosen �42� ways. Thus the probability of a full house is:

𝑃(𝑓𝑢𝑙𝑙 ℎ𝑜𝑢𝑠𝑒) = �13

1 � �43� �

121 � �4

2�

�525 �

= (13)(4 × 3 × 2)(12)(4 × 3)(52 × 51 × 50 × 49 × 48)

= 0.00144

: For this example, we considered the number set { 0 1 2 3 4 5 6 } (a) We calculate the number of ways to make a 3-digit number out of this set (the first digit can’t be zero). For the first digit in the hundreds position, there could be 6 possible numbers (1-6); for the tens position, there would be 6 remaining digits to choose from (this time 0 can be used); and then there would be 5 options for the unit position. So the number of ways to make a 3 digit number is: 6 x 6 x 5 = 180 ways (b) Now we find out how many ways a 3-digit odd number could be made using this set. To be odd, it must end in a 1, 3, or 5. So there are 3 choices for the one’s digit, then 5 choices for the hundred’s digit (since an odd number is already taken, and it can’t be 0 either), and then we have 5 remaining choices for the ten’s digit, since a zero could go there. So the number of ways an odd number can be made is: 5 x 5 x 3 = 75 ways (c) Now we find out how many ways we could make a 3-digit number greater than 330. This would be the case if there were a 4, 5, or 6 in the hundred’s place, and then the other digits could be picked 6 x 5 different ways; giving 3 x 6 x 5 = 90 ways. The number would also be greater than 330 when a 3 was in the hundred’s position, and then a 4, 5, or 6 in the ten’s position, leaving 5 choices for the unit position, giving 1 x 3 x 5 = 15 arrangements. So the total number of ways the set could be arranged to be greater than 330 is equal to: (3 x 6 x 5) + (1 x 3 x 5) = 105 ways

(b) Next we look at the one pair hand. For a one pair, you must have a pair of cards with the same number, and three other cards which do not match one another or the pair you already

have. For the pair, there are �131 � possible numbers, and once selected, �4

2� possible suits.

The three single cards can be chosen �123 � ways, and each card can have �4

1� suit. Multiplying

these factors and dividing by �525 � gives you the probability for a one pair hand.

𝑃(𝑜𝑛𝑒 𝑝𝑎𝑖𝑟) = �13

1 � �42� �

123 � �4

1� �41� �

41�

�525 �

= 0.42

(c) Now we considered a straight, which is a hand with all five cards being in consecutive order, but not the same suit. Aces can be either high or low. We come up with 10 possible consecutive cards: (ace, 2,3,4,5); (2,3,4,5,6); … ; (10, jack, queen, king, ace). Then, for each of

the five cards it could be �41� suit. But 10 x 4 of these are straight flushes, having all the same

suit. Therefore, the probability of getting a straight is:

𝑃(𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡) = 10 × �4

1� �41� �

41� �

41� �

41� − 40

�525 �

= 0.00392

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