Problem 1a

How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?

First you choose the team of two from among one dozen people. Then you chose the team of 5 from the remaining 10 people. You have to divide by a factor of 2 prime which you are overcounting. Point made here.

\[ {12 \choose 2}*{10 \choose 5} \over 2! \]

choose(12,2)*(choose(10,5)/2)
## [1] 8316

Another way to solve this problem is with the multinomial coeficient, we overcounted by a factor of 2 prime

\[ 12! \over 2!5!5!2! \]

factorial(12)/(factorial(2)*factorial(5)*factorial(5)*factorial(2))
## [1] 8316

Problem 1b

How many ways are there to split a dozen people into 3 teams, where each team has 4 people?

First you have to choose one team of four from a dozen people, there are 12 choose 4 possibilities. Then you have to chose another team of four from the remaining 8, 8 chose 4. This could be followed by an implicit 4 choose 4. You overcounted by 3 factorial because there are three teams.

\[ {12 \choose 4}*{8 \choose 4} \over 3! \]

(choose(12,4)*choose(8,4))/factorial(3)
## [1] 5775

This can also be done with the multinomial coeficient, we have a group of twelve people and we want to break them up into three teams of four. We over counted by a factor of 3! since we don’t care which team is which

\[ 12! \over 4!4!4!3! \]

factorial(12)/(factorial(4)*factorial(4)*factorial(4)*factorial(3))
## [1] 5775

Can you visualize these problems with the multiplication principle? Or in a tree diagram?

Problem 2

Consider the identity, what is the appropriate stroy proof for this identity?

\[ {n \choose k} + {n \choose k-1} = {n+1 \choose k} \]

There are 45 ways to choose teams of two (pairs from a group of ten people)

choose(10,2)
## [1] 45

There are ten ways to chose single man teams from a group of 10 people

choose(10,1)
## [1] 10

The sum of thi is 55 different ways.

This is the same as chosing pairs from pairs of two from 11 people.

choose(11,2)
## [1] 55

For this problem, the answer is:

Consider n+1 people, with one of them pre-designated as the “president” of the group. The right-hand side is the number of ways to choose k out ofthese 1 people, with order not mattering. The left-hand side counts the same thing in a different way, by considering two disjoint cases: the president is not in the chosen subset (first term) or is in the chosen subset (second term).

How can I think of this in terms of sets, understand it more intuitively, come up with my own story proof?

What do thy mean by disjoint cases?

Problem 3a

P(the total after rolling 4 fair dice is 21) >/</= P (the total after rolling 4 fair dice is 22)

P(A):

Here are the different rolls and how many possibilites there are to get them:

6663 (4 possibilities) 6654 (12 possibilities) 6555 (4 possibilities)

choose(4,3)
## [1] 4
factorial(4)/(factorial(2)*factorial(1)*factorial(1))
## [1] 12
choose(4,1)
## [1] 4

For P(B), the different possibilities are:

6664 (4 possibiliteis) 6655 (6 possibilities)

choose(4,3)
## [1] 4
choose(4,2)
## [1] 6

Probability of getting a 21:

\[ 20/4^6 \]

Probability of getting a 22:

\[ 10/4^6 \]

20/(4^6)
## [1] 0.004882812
10/(4^6)
## [1] 0.002441406

Problem 3b

A palindrome is an expression such as “A man, a plan, a canal: Panama” that reads the same backwards as forwards, ignoring spaces, capitalization, and punctuation. Assume for this problem that all words of the specified length are equally likely, that there are no spaces or punctuation, and that the alphabet consists of the lowercase letters a,b,…,z.

We can calculate the probability that a two letter word is not a palindrome which would simply be choosing 1 letter and then chosing a different letter, this automatically makes a two-word word not be palindrome. Divide this by the total number of two letter words.

\[ {26*25 \over 26^2} \]

Now, using the subtraction principle we get.

\[ 1- {26*25 \over 26^2} \]

This probabilit is the same as 1/26

1-((26*25)/(26^2))
## [1] 0.03846154
1/26
## [1] 0.03846154

I am having trouble convincing myself that the probability of a random 3 letter word being a palindrome is the same as the probability of a two letter word, the two letter word is easy to calculate (see above)? I need to to calculate the probaiblity that a random three letter word is a palindrome?

Problem 4

This one was seemed to be the right choice given the intuitive definition of naive probability, but I need to explain it to another human and see if I make sense.