Emily Key Davis ST 371-002

Dr. Smith 2/2/2011

Homework Zero: Lecture Summary of 1/27/2011 Turned in HW#1 Continued with Birthday probability problem (from previous class) Data = ½ class with birthday dates collected Handout other “Birthday problem” and HW#2 due 2/3/2011 On HW#2: 2) Adoption 3&4) Van diagram

5) Collectively exausted UAi6&7) Row Charts

=Severything

8) Van diagram Gradebook on wolfware (3.1) “Birthday Problem” from handout Q1-do at least two people have same birthday? A=”at least 2 people same birthday”

A’=”all different” Sample Space (S): (365)P(A)=(# sequences in A)/(365

K K) or P(A)=1-P(Ac

*by 23 people P(A) over .5 ) EASIER!

Q2—probability that someone has same birthday as the 1st

P(B)= 1 – P(B person?

c

P(B)

c

Committees Combination Problem: )=365/365 x 364/365 x 364/365….

Problem 7: Group 5 Women 7 Men

kCnnCkkn

,==

!)!(!

kknn−

!),(, kkCnkPn =

Q1: How many different committees with 2 women and 3 men? (too big to draw a tree) Multiplication Rule 2 Steps:

(1) Males

37

sample size over number choosen

(2) Women

25

350)!2)!25(

!5(!3)!37(

!725

37

=−−

=

xx

Q2: How many different committees if 2 men feuding? (will be <350) Men (how many ways can pick with out feuders:

− 5

37

Women are ok

Could either choose 1 feuding guys or no feuding guys:

+

=

−

25

35

02

25

12

25

537

x

Handout from Drake book—Unranked committees: 4 males (R,S,T,U) and 5 Females (V,W,X,Y,Z) Q: R&S cannot be on same committee unless 1 female member of committee with 4 members:

If no constraints:

49

Possible results:

RS’

37

R’S

37

R’S’

47

RS f1(with one female)

12

15

F2 (with 2 females)

02

25

Result: 12525

12

15

47

37

2 =

+

+

+

MUCH EASIER!!!! P(A)=1-P(A’)

i) broke into disjointed events ii) solve each event iii) sum up all possible possibilities from all events (ADD)

OR look at complement

125112644

05

49

=−=

−

p. 12: Example 10: Combination Problem Box w/ candy 3 colors: 5 blue, 6 green, 7 red randomly pick 3 Q1: What is probability all three same color? (Addition Rule) A= all 3 same color A=all red + all blue + all green

6536

05

07

06

35

07

06

05

37

=

+

+

=A

Pr(A)=65/816 Q2: What is Probability that one ball of each color is selected? (MULTIPLICATION RULE)

21016

15

17

=

=B

Pr(B)=210/816 Example 11: Q: We want to arrange 8 letters (A,A,B,B,C,C,D,D) in a row POSITION MATTERS 1)How many ways can we arrange the letters? *but A=A B=B C=C D=D INDISTINGUISHABLE Example done with TOOT = 4! / (2!2!) A = arrange 8 letters = 2! * 2! * 2! * 2! correct for indistinguishable

8! (if distinguishable)

*Question asked: Why indistinguishable—explained using TOOT example—collapsing branches of tree to give same result as math

Handout: Problem 7

a) STATISTICS breaking objects into indistinguishable groups Letter multiplicity notation

54321

12133n

C

n

I

n

A

n

T

n

S

Permutations = 504000!1!*2!*1!*3!*3

!10!!*!*!*!*

!

54321

==nnnnn

n

Problem 8: 9 people in 3 cars: All cars must go! C1holds 2 C2holds 4 C3 Notation:

holds 5

321 ###9

CCC if C1+C2+C3

But does not = 9 because 2+4+5=11 now must take 3 cars still!

= 9

Drew Tree on board # arrangements =

4410522

9432

9342

9531

9441

9###

9

321

=

+

+

+

+

=

∑

CCCBecause do not care what position in cars people are sitting!

Taner Howard Lecture Date: 1-27-2011 The lecture started off by giving a preview of the second homework assignment. He stated that the first problem involving arrangements of repeated letters would be discussed in this lecture. There was a note made about number two that one could only adopt a rule if he or she was not already following it. Then, it was mentioned that the other problems deal with making Venn diagrams or combinatorics. Also, a reminder was given that there are also two problems to work out in Devore. The next topic in the lecture was to analyze the birthday problem which had been started the previous lecture. First, we considered the probability that any two people in the room had the same birthday. It was said that it would be easier to find the probability of the complement, the probability that everyone does not have the same birthday. The sample space was 365 days of the year raised to the n power where n is the number of people in the class. The first person in the class could be any of 365 days; the second could be any of 364, and so on. Therefore, the number of ways the complement could occur would be (365!)/(365-n)!. If this number of ways is divided by the sample space, we have the probability of the complement. To find the probability of any two people sharing the same birthday, the probability of the complement had to be subtracted from one. This probability came out to be fairly higher than expected. Next Dr. Smith explained why the probability had been higher than we had expected it to be. A lot of us were thinking about the case that someone would share a birthday with his or herself. In fact this probability was a lot lower than the first case. This probability was calculated in the same fashion as above except the probability of the complement was calculated in a different way. Now the number of ways the complement could occur was 365(364(n-1)). This was because your birthday could be any of the 365 days of the year but, everyone else’s birthdays could be any of the other 364 days. There was a handout given in class about the birthday problem as well. Before moving on to the next topic, there was a brief mention about notation and the relationship between combinations and permutations. It was mentioned that a combination could be expressed in a binomial coefficient notation, a nCk notation, a Cn,k notation, or just the n!/((n-k)!k!) definition. The other simple, yet important, relationship given was that a combination of n choose k things multiplied by k factorial equals the permutation of those n things taken k at a time. The next topic of discussion was about committee combinations. The first example was a problem done on the board. There are five women and seven men. From these, we want to form a committee consisting of two women and three men where order does not matter. To find out how many different ways this is possible, we use the multiplication rule and combinations. To find the answer we multiply the number of combinations of three men from the seven by the number of combinations of two females taken from the five. The second part of the question asked how many ways could the same committee be formed if two of the men refused to serve together? To answer this question, multiplication rule and addition rule were used. The multiplication rule is used in the same way as above; it is used to multiply the number of combinations of women by the number of combinations of the men. The only difference is that the addition rule is used to find the number of combinations of men that are possible for this scenario. Specifically, the addition rule adds the

combination of choosing three men from the non-refusing men to the combinations of choosing two men from the non-refusing men and one of the refusing men. Next there was a handout given about another committee problem. It was very similar to the above problem. However, one important twist that does need to be pointed out is the fact that using a tree diagram for part of a combinatorics problem can be extremely helpful. At the end of each branch a mutually exclusive combination can be calculated. Then, the number of combinations can be added. Following the committee combinations, we discussed different probabilities of drawing different combinations of colored balls from a box. The box contains five blue balls, six green balls, and seven red balls. First, we considered the probability of drawing three balls of the same color. To find this, the addition rule was used to add the probability of drawing three reds, three greens, and three blues. The second case was finding the probability of drawing three balls all of different colors. To find this, the multiplication rule was used by multiplying the probability of drawing one green, one blue, and one red. The last major topic of the class was finding out how many different ways to arrange letters when there were repeating letters. A problem was worked on the board on this topic and, there was a handout given on the number of ways “statistics” could be arranged. The basic principle of working these kinds of problems is to first find the number of ways the word could be arranged if the repeating letters were distinguishable and then divide by the number of times each letter was repeated. Also, on this last handout, there was a problem about arranging nine people in three cars.

Homework 0 1/27 by:Matt Green

Today in Stat 370 we talked about combinations and permutations again. We started out talking about the birthday problem. In this problem we are looking for the chances that two people out a selected group will have the same birthday. The outcome was surprising. The best way of thinking about the chances is to look at the situations in which no one has the same birthday. This is shown in the equations of 365/365 * 364/365 * 363/365 * … * (365-n)/365.As you can see in order for two people to not have the same birthday each time the nominator has to decrease by one. When you get to n(the number of people in your group), you will see the chances that two people will not have the same birthday. If you wanna know the chances that two people will have to same birthday, you just take the number you found and subtract it from one. The next thing we look at was notation. (n/k) =nCk=Cn,k all mean the same thing which is n!/(n-k)!k!. This means you have n object and you are picking k out of them. We went over a example in class about a committee being formed out of 7 men and 5 woman. The committee is formed by 3 men and 2 women. So you do (7/3) for the men and (5/2), which turns into 7!/(4!)3! * 5!/(3!)2! The next example we did was about drawing marbles out of a bag. We were looking for the different ways you could draw out certain situations. The first way we looked for was when you pulled out balls of all the same color. If there were 7 red balls 5 blue balls and 6 green balls, then the chances would be ((7/3)+(5/3)+(6/3))/(18/3). The last type of example we did was rearranging words, this is normally easy because each word we have looked at thus far has had individual distinguishable letters. A word like dog is easy figured out by taking the amount of letters and putting a factorial after it. So dog would be 3!. This becomes a little more complicated when you get a word like good because there are 2 o's in it. The way you deal with this is pretend that the o's are separate , and that makes it 4! but because there were 2 o's you put it over 2!, so your result is 4!/2!. For words like, toot, it would be 4!/2!2!

Lecture Summary for Statistics 371 Class Date: 1/27/2011

By Shawn Humble Review of Definition of Specific Terms used in Probability Collectively exhaustive means the entire sample space. Ai ∩Aj are ∅if disjoint or mutually exclusive (disjoint and mutually exclusive are synonyms). Birthday Problem First, we divided the classroom into two groups of about 40 students each. We then asked what are the chances that two people in a group have the same birthday. (ignoring Feb 29) If we consider that it is equally likely that a person can be born on any given day, than for each person there are 365 possible dates for each person. For a group of N students (N=40) in our case, there are 36540 possible outcomes or 3.1 x 10102 outcomes (higher than many calculators can do without coming up with an error). Now, we’d have to count up all the possible outcomes that could occur where two people have the same birthday and then divide it by 3.1 x 10102 This is difficult to do so we need to look at it a different way. If Event A = {At least two people have the same birthday} Then Event AC = {All N people have different birthdays} (40 in our case) This becomes a permutation problem since if the first person uses up one birthday, there are only 364 for the next person to choose from, and then 363 for the next person, etc., etc.

So, )!40365(

!365−

=CA = 3.37 x 10101 cases where all N people have different birthdays (permutations). So there are 2.76 x 10102

situations (permutations) that have the same birthdays in them. So the chances of all 40 people having different birthdays are (AC/36540)=3.37 x 10101/3.1 x 10102 which comes to about a 10.1% This means that P(A)=1- P(AC) =1-10.1% = 89.1% So in a group of 40 people there is a very high chance that two people will have matching birthdays. Difference between Combinations and Permutations

Combinations: The number of possibilities when order does not matter. (i.e., {1,2,3} is the same as {3,2,1}) Permutations: The number of possibilities when order does matter (i.e., {1,2,3} is different than {3,2,1})

Combinations are represented for us by which is nkk CnCkn

,≡≡⎟⎟⎠

⎞⎜⎜⎝

⎛)!()!(

!, kkn

nC nk −=

Permutations by which is nkP , )!(!

, knnP nk −

=

The letter n represents the number of objects in the group that we are choosing from, and the letter k represents the actual number of objects that we will choose. So if there is a group of 5 men and you want to choose 2 men from the group, then you would have n=5 and k=2.

Problem of Committees If you have a group of 7 men and 5 women, how many different committees could you make if there were 2 women and 3 men in each committee. Case I: If you didn’t care about what the order would be and just needed 5 people, this is a combination type problem. Let’s first consider if we didn’t care whether the group of 5 people was men or women.

You would then have possible combinations which is 792 possible committees. ⎟⎟⎠

⎞⎜⎜⎝

⎛5

12

But we want to fix this to a certain number of men and women so we will have less than this amount.

For 3 men, the most combinations we can make out of the 7 men would be or 35 different combinations. Now for each of these

21 combinations, we have or 10 combinations. So by the product rule, we have 35*10=350 possible committees.

⎟⎟⎠

⎞⎜⎜⎝

⎛37

⎟⎟⎠

⎞⎜⎜⎝

⎛25

This is less than the 792 possible committees, so that is a good sign. Case II: What if 2 men refuse to be on the same committee? This means we will have even less possibilities.

So now for the men, we have either or combinations. The first is when we have a combination with at least one of

the men in the committee and 2 other men not in the committee and the second is when we have a committee without either of the two men. Since these are disjoint, by the addition rule, we add these two values together to get the total number of combinations for the men.

⎟⎟⎠

⎞⎜⎜⎝

⎛12

⎟⎟⎠

⎞⎜⎜⎝

⎛25

⎟⎟⎠

⎞⎜⎜⎝

⎛02

⎟⎟⎠

⎞⎜⎜⎝

⎛35

⎟⎟⎠

⎞⎜⎜⎝

⎛12

⎟⎟⎠

⎞⎜⎜⎝

⎛25

=20 and =10 so we have 30 possible combinations for the men. The women stay the same so for each man

combination, we have 10 possible women combinations. Our total value is now 300 possible committees.

⎟⎟⎠

⎞⎜⎜⎝

⎛02

⎟⎟⎠

⎞⎜⎜⎝

⎛35

This is less than our original 350 so while we don’t know if we are correct, it is a good sign. Another Example: Problem of Committees From the handout in class, we have the unranked committee problem. We have 4 males and 5 females. We want to form a committee of four members, and we are given the constraint that 2 of the males cannot be on the same committee unless there is at least one female on the committee. How many committees can be formed?

If we have no constraints, we have possible combinations which are 126 possible committees. ⎟⎟⎠

⎞⎜⎜⎝

⎛49

Solution I: One way to look at this, is how many possible combinations are there that cannot result in a possible combination. This

will happen when we choose both fighting males, and no females. So we would have =1. So, out of the two fighting

men, we choose both of them, only one way to do that, for the non-fighting men, we choose both of them, only one way for that, and we don’t choose any women so only 1 way to do that.

⎟⎟⎠

⎞⎜⎜⎝

⎛22

⎟⎟⎠

⎞⎜⎜⎝

⎛22

⎟⎟⎠

⎞⎜⎜⎝

⎛05

So possible committees would be 126 total combinations minus the 1 combination that we don’t allow for a total of 125 committees.

Solution II: Another way to do this is to draw a tree with all of our possible combinations and then add them together.

RS’* =35 combinations ⎟⎟⎠

⎞⎜⎜⎝

⎛37

R’S* =35 combinations ⎟⎟⎠

⎞⎜⎜⎝

⎛37

R’S’* =35 combinations ⎟⎟⎠

⎞⎜⎜⎝

⎛47

⎟⎟⎠

⎞⎜⎜⎝

⎛12

⎟⎟⎠

⎞⎜⎜⎝

⎛15

= 10 combinations

RS = 10 combinations ⎟⎟⎠

⎞⎜⎜⎝

⎛02

⎟⎟⎠

⎞⎜⎜⎝

⎛25

⎟⎟⎠

⎞⎜⎜⎝

⎛22

⎟⎟⎠

⎞⎜⎜⎝

⎛05

(event not allowed since R and S are both in the group, we must have one woman)

If we add up all the endings of our branches, we get 35+35+35+10+10 which equals 125 combinations. This matches our solution from Solution 1. It is important to realize that each branch ending is has the R and S multiplied to the rest of the combinations. For example:

RS’ stands for =1 possible combination, so we are still satisfying the Product Rule. ⎟⎟⎠

⎞⎜⎜⎝

⎛11

⎟⎟⎠

⎞⎜⎜⎝

⎛01

Box Example We now move to a box that is filled with 3 different colored balls. We have 5 blue balls, 7 red balls, and 6 green balls. Without any constraints, what is the number of ways that we can choose 3 balls from the box?

We have a total of 18 balls and we will choose 3 so possible combinations (order doesn’t matter) are: = 816 combinations. ⎟⎟⎠

⎞⎜⎜⎝

⎛3

18

Case I: Event A = {all 3 balls are the same color} What is the probability that Event A will happen? First, what are the number of combinations where all the balls can be blue?

Blue = which equals 10 possible combinations. ⎟⎟⎠

⎞⎜⎜⎝

⎛35

⎟⎟⎠

⎞⎜⎜⎝

⎛06

⎟⎟⎠

⎞⎜⎜⎝

⎛07

Red = which equals 35 possible combinations. ⎟⎟⎠

⎞⎜⎜⎝

⎛05

⎟⎟⎠

⎞⎜⎜⎝

⎛06

⎟⎟⎠

⎞⎜⎜⎝

⎛37

Green = which equals 20 possible combinations. ⎟⎟⎠

⎞⎜⎜⎝

⎛05

⎟⎟⎠

⎞⎜⎜⎝

⎛36

⎟⎟⎠

⎞⎜⎜⎝

⎛07

Since these are all disjoint, we have 65 possible combinations that occur when all three balls will be the same color. The probability of this happening is 65/816 which is about an 8% chance.

Case II: Event B={One of each ball is selected}. What is the probability that this will happen? The only possible way this can happen is with the following combinations:

⎟⎟⎠

⎞⎜⎜⎝

⎛15

=210 possible combinations when each ball is a different color. ⎟⎟⎠

⎞⎜⎜⎝

⎛16

⎟⎟⎠

⎞⎜⎜⎝

⎛17

The probability for this occurring would then be 210/816 which is about a 26% chance of happening. So if you were gambling, you would be better off guessing that each ball will be different than that each ball would be the same. Letter Arranger Example Let’s say that we have the letters {A,A,B,B,C,C,D,D} If all the letters were different how many combinations would be possible? Let’s start with an easier example: Dog = {D,o,g; D,g,o; o,D,g; o,g,D; g,o,D; g,D,o} for six possible combinations or 3! combinations. The word good (if the two o’s were indistinguishable would be) 4! combinations or 24 combinations. But, since the two letter o’s are the same, we will have 4 (or 2!) possible ways that each of these can be arranged so we need to divide the total number of combinations by 2!. So 4!/2! = 6 possible combinations. Now, back to the original problem.

We have the letters {A,A,B,B,C,C,D,D} So we have 8 total letters. But we have 4 combinations where the letters are the same so we need to divide this out of the total number of combinations.

!2!*2!*2!*2!8 =2520 ways to arrange the letters above.

Another Letter Arranger Example Same problem above, but with the word STATISTICS Number of letters (n=10) Number of different combinations of the same letter: 3(S), 3(T), 2(I) So, we need to divide out these multiple combinations.

This gives us !2!*3!*3

!10 =50,400 ways to arrange the letters above.

People in a Car Example 9 people need to fit in 3 cars, and each car MUST contain at least 1 person. Car 1 can fit 2 people, Car 2 can fit 4 people, and Car 3 can fit 5 people. So, if we take the case where car 1 only has 1 person in it, we have only two options for the other 8 people. Either 4 go in Car 2 (total amount) and 4 go in Car 3. Or we have 5 in Car 3 (total amount) and the other 3 in Car 2.

These possible combinations are: !4!*4!*1

!9 or !3!*5!*1

!9

You add these combinations together, and then you determine how many combinations there would be if there were 2 people in Car 1, and the other 7 spread through Cars 2 and Car 3.

You would find that the possible combinations are: !3!*4!*2

!9 or !4!*3!*2

!9 or !5!*2!*2

!9

Since all these events are disjointed, we can add them all up using the addition rule to find that there are 4410 different possibilities for dividing these 9 people between the 3 cars.

ST371-002 CLASS SUMMARY

January 27, 2011

Melody Wilson

Class started out with reviewing the birthday problem from Tuesday’s class (1/25). Only one side of the class had two people with the same birthday. It was kind of what we had expected.

Homework 2 was assigned and due the next Thursday (2/3) (now changed to Tuesday the 8th

A handout was passed out with a sample problem that went into more detail about the birthday problem. For example, A

). Smith talked through some of the problems. He explained that collectively exhaustive is everything in the sample space, and disjointed is another way of saying mutually exclusive, or the null.

C is all k people having different birthdays, so it asked for the P (AC

365365

× 364365

× 363365

× … × (365−(𝑘−1))665

):

A new problem was proposed, where there are 5 women, and 7 men in a group, and Smith brought up the concept of combination:

�𝑛𝑘� = 𝐶𝑘,𝑛 = 𝑛!(𝑛−𝑘)!

Therefore, how many different committees could have 2 women and 3 men? The solution is as follows:

�52� = 5!(5−2)!

and �73� = 7!(7−3)!

Now using the multiplication rule:

�52� × �73� = 350 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒𝑠

There is also the case where 2 of the men are feuding and refuse to serve on the committee, so how many different committees can be formed now still with 2 women and 3 men?

�52� × �73 − 5� = ��21��52� + �20��

53�� × �52�

Where, 5 = �22� × �51�

Since the above concept was tricky, another example was presented. This was using event space

. This committee has 4 members formed from a group of 4 males denoted R, S, T, U and 5 females denoted V, W, X, Y, Z. R and S cannot be on the same team unless there is one female. The solution is as follows:

Then, add the above together, which gives 125 committees. This could have also been done using the complement. There are, without restrictions, 126 combinations:

�94� = 126 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 − ��50��44�� = 125 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠

The next example was a probability question with a box that contained 5 blue balls, 6 green, and 7 red balls. 3 balls were randomly selected. With no restraints the solution is:

�183� = 816

Now, what is the probability that all 3 balls have the same color? Use the addition rule to solve (A = sample space):

𝐴 = # 𝑎𝑙𝑙 𝑟𝑒𝑑 + # 𝑎𝑙𝑙 𝑏𝑙𝑢𝑒 + # 𝑎𝑙𝑙 𝑔𝑟𝑒𝑒𝑛

𝐴 = �73� �

50� �

60� + �

53� �

60� �

70� + �

63� �

50� �

70� = 65

So, the probability would be:

𝑃(𝐴) = 65

816

What is the probability that one ball of each color is selected?

𝐵 = �51� �

61� �

71� = 210

𝑃(𝐵) = 210816

For the next example, the order matters! There are 8 letters {A, A; B, B; C, C; D, D} and we want to arrange them in a row. How many ways can we arrange them if each letter is distinguishable?

8!2! × 2! × 2! × 2!

= 𝑤𝑎𝑦𝑠 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 (𝑑𝑖𝑠𝑡𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒)

8 is the number of letters in all, and each 2 is how many of each letter there are, e.g. there are 2 A’s, 2 B’s, etc.

Andrew Morgan

Date of class: 1/31/11

At the beginning of the class the results of the previous experiment were discussed. The

previous experiment was to see if two people in the class had the same birthdates. The results

were that there was one pair of people who had the same birthday, which was about average or

below average for a class our size. The birthday problem uses the pigeonhole principle to show

that when the sample of people reaches 367 there is a 100% chance of two people having the

same birthday and about 50% with 23 people.

Statistics 371 Class Summary

After going over the homework that will be due the following week, the class started

going back over the topic that started last time, counting. For this class, only combinational

counting problems were covered. We defined what basic combinations were. They followed the

basic equation of 𝐶𝑘,𝑛 where you have “n” objects and you pick “k” of them.

The first problem we did was example 7 that was handed out in class. The question asked

how many permutations of letters in the word “statistics” existed. To solve this, you have to first

assume max permutations as if each letter was distinct, in this case it would be 10!. Then you

have to divide out the duplicate letters. The final answer for this problem was:

10!

3!∗3!∗2!

We did another example of a combinational counting problem that involved groups of

men and women. The problem states that there are 5 women and 7 men. How many different

committees consisting of 2 women or 3 men can be formed? We found that if you use the

combination formula for each case and multiply them, you will get the answer. So in this

problem you have 7 males, and you are choosing 3, and you have 5 females and you are

choosing 2. The following equation represents this.

�73� ∗ �5

2� = � 7!

4!∗3!� ∗ � 5!

(3!∗2!)� = 350

We then expanded on this problem and said what if two guys couldn’t sit next to each

other? There are two solutions to this problem. You could either subtract the cases you didn’t

want or you could add every acceptable male case together and multiple it with the females like

normal.

Next, we had another example on combinations where we had a box filled with different

colored balls (5 blue, 6 green, and 7 red). The problems were 1.) What is the probability that all 3

balls will be the same color and 2.) What is the probability that 1 of each color will be selected.

For the first question you use the combination formula where you take three of each color and

add them together. For the second problem you just multiple 3 combinations together where you

take one of each color.

To end the class we reviewed another combination problem with cars and did another

letter permutation problem. The car problem was passed out with number 7 mentioned at the

beginning of the summary.

Jonathan Gregory ST 371-002

Summary of 1/27/11

The objective of this class was to learn about and solve permutations of probability problems. The first example we went over was the “Birthday Problem”. The question asked was, “Out of a group of k randomly selected individuals, what is the probability that at least two will have the same birthday? (omit February 29 birthdays)”. We decided to set our variable:

𝐴 = At least two people have the same birthday We noticed our sample space was (365)k

So we decided to name another variable use it to solve the question: 𝐴𝑐 = All k people have different birthdays

𝑃(𝐴𝑐) =365365

∗364365

∗363365

∗ … ∗365 − (𝑘 − 1)

365

𝑃(𝐴) = 1 − 𝑃(𝐴𝑐)

which could get awfully large considering:

𝑃(𝐴) =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒𝑠 𝑖𝑛 𝐴

(365)𝑘

Our next question was similar: “What’s the probability that at least one other person has the same birthday as the FIRST person?” We used the same procedure.

𝑃(𝐵𝑐) = 365365

∗364365

∗ … ∗364365

𝑃(𝐵) = 1 − 𝑃(𝐵𝑐) Our next example involved a committee of 4 members that was to be picked from a selection of 4 males (R, S, T, U) and 5 females (V, W, X, Y, Z). It was specified that R and S cannot be on the committee together unless one female is also on the committee. We started by finding decomposing the problem into smaller pieces until we were left with nothing but mutually exclusive events shown in the graph below.

𝑅𝑆′ = �73�

𝑅′𝑆 = �73�

𝑅′𝑆′ = �74�

𝑅𝑆 = �73�

𝑓1 = �51� ∗ �

21�

𝑓2 = �52�

Event X: X is on the committee (X’ meaning X is not on the committee) Event fn: Exactly n females are on the committee The top numbers (t) represent the total number of people to select from and bottom number (b)

represents how many are going to be picked using the formula 𝑡!(𝑡−𝑏)!𝑏!

.

Adding the mutually exclusive events together we learned that there were 125 possible combinations.

Jonathan Gregory ST 371-002

Summary of 1/27/11 The next topic we worked on involved distinct permutations of letters in words. Our first example used the word “STATISTICS”. We divided the word into individual, unique letters. Letter S T A I C Multiplicity 3 3 1 2 1 Notation n1 n2 n3 n4 n5 We found the number of permutations using the following formula.

𝑛!𝑛1!𝑛2!𝑛3!𝑛4!𝑛5!

=10!

3! 3! 1! 2! 1!= 50,400

Our last example involved the number of distinct permutations of 9 people fitting into 3 cars.

The cars were labeled C1, C2, & C3. C1 could hold 2 people, C2 could hold 4, and C3 could hold 5. We knew the total amount of permutations was equal to:

�9

#𝐶1, #𝐶2, #𝐶3�

where #C1+#C2+#C3 = 9. We then listed the possible combinations of people per car.

#C1 #C2 #C3 1 4 4 1 3 5 2 4 3 2 3 4 2 2 5

We then found the number of possible combinations.

�9

#𝐶1, #𝐶2, #𝐶3� = �

91,4,4

� + �9

1,3,5� + �

92,4,3

� + �9

2,3,4� + �

92,2,5

� = 4,410