Help with a Markovian chain, Liner programming and Probabili

[inso]

I’m preparing for a final and I’ve stumbled upon the following problems that I just can’t get right. Any help with them will be extremely appreciated!

Markovian chain problems
1. A carnival man moves a pea among three shelves, A, B and C. Whenever the pea is under A, he moves it with equal probability to A or B( but not C) When it is under B, he is sure to move it to C. When the pea is under C, it is equally Likely to move it to A, B or C.
1. Set up a three state Markov chain by drawing the transition diagram and check whether it is regular
2. Find the transition matrix and evaluate the probabilities for the pea to be under each shell after two moves, if it is initially under C
3. Find the long range probabilities for the pea to be under each shell
4. Find the probabilities for the pea to be under each shell after two moves, if you know that it was initially under C and the next time it is not under B
2. A professor tries not to be late for class too often. If he is late some day, he is 90% sure to be on time the next time. If he is on time, the next day there is a 25% chance of his being late. In the long run, how often is he late for class?

Linear programming
A farmer owns a 200-acre farm and can plant any combination of two crops 1 and 2. Crop 1 requires 1 man-day of labor and $10 of capital per each acre planted, while crop 2 requires 4 man-days of labor and $20 capital per acre. The farmer has $2200 of capital and 320-man-days of labor available for the year. Crop 1 produces $40 of net revenue and Crop 2 produces $60.
a) Find the optimal strategy
b) Because of some tricks on the stock-market the price of Crop 2 begins growing up. What is the turning point of the price of Crop 2, when the farmer should recalculate the maximum revenue scheme?

Probabilities
1. Find the probability for the following poker hands:
a) Full house which contains at most 2 Face cards
b) Three of a kind, given that it contains exactly one club
c) Four of a kind, given that is contains at least 3 red cards

2. An urn contains 2 Black and 4 Red balls. A sample of 2 balls is drawn and then withought replacement two more balls are pulled out
a) Check whether the events:
P: The balls of the first sample have the same colour
And
Q: The balls of the second draw have the same colour
Are independent?
b)Find the probabilities for each of the following:
A: The balls of the first draw have the same color, given that both balls of the second sample are red
B: The balls of the first draw have different color, given that the balls of the second one have different colour.

 

[soroban]

Hello, inso!

Markovian chain problem

2. A professor tries not to be late for class too often.
If he is late some day, he is 90% sure to be on time the next time.
If he is on time, the next day there is a 25% chance of his being late.
In the long run, how often is he late for class?

Let \(\displaystyle P\) = he is punctual.
Let \(\displaystyle L\) = he is late.

\(\displaystyle \text{We have: }\;\begin{array}{ccc} & P & L \\ P & 0.75 & 0.25 \\ L & 0.9 & 0.1 \end{array}\)


\(\displaystyle \text{We have: }\;(p,q)\text{, where: }p+q\:=\:1\;\;[1]\)

\(\displaystyle \text{and: }\;(p,q)\begin{pmatrix}0.75&0.25\\0.9&01\end{pmatrix} \;=\;(p,q) \quad\Rightarrow\quad \begin{array}{cccc}0.75p + 0.9q &=& p & [2] \\ 0.25p + 0.1q &=& q & [3] \end{array}\)


\(\displaystyle \text{From [1] and [3], we have: }\;\begin{array}{ccc} p + q &=& 1 \\ \frac{1}{4}p - \frac{9}{10}q &=& 0 \end{array}\)

\(\displaystyle \text{Solve the system: }\;(p,\,q) \;=\;\left(\tfrac{18}{23},\;\tfrac{5}{23}\right)\)

. . \(\displaystyle \text{Therefore, He will be late }\frac{5}{23}\text{ of the time.}\)

 

[inso]

10Q so very much!! Its a life saver...

 

[soroban]

Hello again, inso!

Here's another one . . .

1. Find the probability for the following poker hands

\(\displaystyle \text{There are: }\:{52\choose5} \:=\:2,\!598,\!960\text{ possible 5-card poker hands.}\)

\(\displaystyle \text{This is the denominator }(D)\text{ for all the probabilities.}\)

a) Full House which contains at most 2 Face cards.

A Full House consists of a Triple and a Pair.
. . It can have 2 Face cards or no Face cards.


\(\displaystyle \text{(1) The Full House has 2 Face cards; they must comprise the Pair.}\)
\(\displaystyle \text{There are 3 choices for the value of the Pair and }{4\choose2} \:=\:6\text{ ways to get the Pair.}\)
\(\displaystyle \text{Then there are 10 choices for the value of the Triple (it must not be a Face card)}\)
. . \(\displaystyle \text{and }{4\choose3} = 3\text{ ways to get the Triple.}\)
\(\displaystyle \text{There are: }3\cdot6\cdot10\cdot3 \:=\:540\text{ Full Houses with two Face cards.}\)


\(\displaystyle \text{(2) The Full House has no Face cards.}\)
\(\displaystyle \text{There are 10 choices for the value of the Triple, and }{4\choose3} = 4\text{ ways to get the Triple.}\)
\(\displaystyle \text{There are 9 choices for the value of the Pair, and }{4\choose2} = 6\text{ ways to get the Pair.}\)
\(\displaystyle \text{There are: }\:10\cdot4\cdot9\cdot6 \:=\:2160\text{ Full Houses with no Face cards.}\)


\(\displaystyle \text{Hence, there ae: }\;540 + 2160 \:=\: 2700\text{ Full Houses with at most 2 Face cards.}\)


\(\displaystyle \text{Answer: }\;\frac{2700}{D}\)

b) Three of a kind, given that it contains exactly one Club.

\(\displaystyle \text{A Three-of-a-kind consists of a Triple and two Others (unmatched cards).}\)
\(\displaystyle \text{The one Club can be among the Triple or among the Others.}\)


\(\displaystyle \text{(1) The Club is among the Triple.}\)

\(\displaystyle \text{There are 13 choice for the value of the Triple}\)
. . \(\displaystyle \text{and 3 ways to get the Triple and include a Club.}\)

\(\displaystyle \text{The two Others must not be a Club and they must not match each other.}\)
\(\displaystyle \text{The 4th card can be any of the 39 non-Clubs.}\)
\(\displaystyle \text{The 5th card can be any of the 36 nonmatching non-Clubs.}\)
. . \(\displaystyle \text{There are: }\frac{39\cdot36}{2} =702\text{ ways to get the two Others.}\)

\(\displaystyle \text{There are: }\:13\cdot3\cdot702 \:=\:27,\!378\text{ ways to get a 3-of-a-kind with a Club among the Triple.}\)


\(\displaystyle \text{(2) The Club is among the Others.}\)

\(\displaystyle \text{There are 13 choices for the value of the Triple}\)
. . \(\displaystyle \text{ and }one\text{ way to get the Triple (and }not\text{ include a Club).}\)

\(\displaystyle \text{Let's say the 4th card is a Club.}\)
\(\displaystyle \text{There are 12 Clubs to choose from (we don't want Four-of-a-kind).}\)

\(\displaystyle \text{For the 5th card, there are 48 cards to choose from.}\)
. . \(\displaystyle \text{But one of them matches the Triple.}\)
. . \(\displaystyle \text{Tweve of them are Clubs.}\)
. . \(\displaystyle \text{And three of them match the value of the 4th card (we don't want a Pair).}\)
\(\displaystyle \text{So there are 32 choices for the 5th card.}\)

. . \(\displaystyle \text{There are: }\frac{12\cdot32}{2} = 192\text{ ways to get the 2 Others (with a Club.)}\)

\(\displaystyle \text{Hence, there are: }\:13\cdot1\cdot12\cdot192\:=\:29,\!952\text{ ways to get 3-of-a-kind with a Club among the Others.}\)


\(\displaystyle \text{Therefore, there are: }\:27,\!378 + 29,\!952 \:=\: 57,\!330\text{ ways to get 3-of-a-kind with exactly one Club.}\)


\(\displaystyle \text{Answer: }\;\frac{57,330}{D}\)

c) Four of a kind, given that is contains at least 3 red cards.

\(\displaystyle \text{A 4-of-a-kind cannot have 4 red cards or 5 red cards. }\:\text{(Think about it.)}\)
\(\displaystyle \text{It must have }exactly\text{ 3 red cards: }\:RRRBB\)
. . \(\displaystyle \text{where }RRBB\text{ is the 4-of-a-kind and the 5th card is red.}\)

\(\displaystyle \text{There are 13 choices for the value of the 4-of-a-kind.}\)

\(\displaystyle \text{The 5th card can be any of the remaining 24 red cards.}\)

\(\displaystyle \text{There are: }\:13\cdot24 \:=\:312\text{ 4-of-a-kinds with at least 3 red cards.}\)

\(\displaystyle \text{Answer: }\:\frac{312}{D}\)


Whew! I need a nap!

 

[soroban]

Hello again, inso!

Here's most of the first one . . .

A carnival man moves a pea among three shells: A, B and C.
Whenever the pea is under A, he moves it with equal probability to A or B (but not C).
When it is under B, he is sure to move it to C.
When the pea is under C, it is equally Likely to move it to A, B or C.

1. Set up a three-state Markov chain by drawing the transition diagram
and check whether it is regular.

\(\displaystyle A \;=\;\begin{pmatrix}\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{pmatrix}\)

2. Find the transition matrix and evaluate the probabilities for the pea
to be under each shell after two moves, if it is initially under C.

\(\displaystyle A^2 \;=\;\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0\\0&0&1\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{pmatrix}\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0 \\ 0&0&1\\ \frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{pmatrix} \;=\;\begin{pmatrix}\frac{1}{4}&\frac{1}{4}&\frac{1}{2} \\ \\[-3mm] \frac{1}{3}&\frac{1}{3}&\frac{1}{3} \\ \\[-3mm] \frac{5}{18}&\frac{5}{18}&\frac{4}{9} \end{pmatrix}\)


\(\displaystyle \text{"Initially under C" means the initial vector is: }\;\left(0\;\;0\;\;1\right)\)


\(\displaystyle \text{Therefore: }\;(0\;\;0\;\;1)\,\begin{pmatrix}\frac{1}{4}&\frac{1}{4}&\frac{1}{2} \\ \\[-3mm]\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\ \\[-3mm] \frac{5}{18}&\frac{5}{18} & \frac{4}{9}\end{pmatrix} \;=\; \left(\frac{5}{18}\;\;\frac{5}{18}\;\;\frac{4}{9}\right)\)

3. Find the long range probabilities for the pea to be under each shell,

\(\displaystyle \text{We have: }\;(p\;\;q\;\;r)\;\text{ where }\+q+r\:=\:1\)

. . \(\displaystyle \text{And: }\;(p\;\;q\;\;r)\begin{pmatrix}\frac{1}{2}& \frac{1}{2}&0 \\ 0&0&1 \\ \frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{pmatrix} \;=\;(p\;\;q\;\;r)\)


\(\displaystyle \text{We have the system: }\;\begin{array}{ccc}\frac{1}{2}p \quad + \frac{1}{3}r &=& p \\ \quad\;\; q + \frac{1}{3}r &=& r \\ p +q + r &=& 1 \end{array}\)

. . \(\displaystyle \text{which has the solution: }\p\;\;q\;\;r) \;=\;\left(\frac{3}{13}\;\;\frac{4}{13}\;\;\frac{6}{13}\right)\)