1202 probabiility help please!!!! (1 Viewer)

aanthnnyyy · Apr 28, 2016

Upcoming tutorial quiz coming up nxt week and probability still killing me from high school to uni

could anyone please help me with these few Q's, textbook doesnt even display the answer. Please do note that im not just seeking solutions, i really want to know how to tackle these problems

1. How many ways can letters of the word SPECIAL be arranged such that the letters SPE appear together and in that order

2. A school has 500 pupils. A pupil who is absent has a probability of 0.2 of returning to school tomorrow. A pupil who is not absent today has a probability of 0.6 of being absent tomorrow. If 324 pupils are absent today, how many will be absent 3 days from now

3. Josh plays a game with three other friends. To win, Josh has to roll the die with a value higher than each his friends'. Josh's friends have values 7, 11 and 2 respectively. If Josh rolls a 20-sided die three times, what is the probability that he wins against all three friends

aanthnnyyy · Apr 28, 2016

:bump:

parad0xica · Apr 28, 2016

1. Consider the string "SPE" as a unit. The set of elements we need to arrange now is {"SPE", "C", "I", "A", "L"} and this can be done in 5! ways.

There are 5 positions: _ _ _ _ _.

Pick up "SPE", how many places can you put this? 5.

Now we have _ _ SPE _ _ .

Pick up "C", how many places can you put this? 4.

Now we have _ _ SPE _ C.

Pick up "I", how many places can you put this? 3.

Now we have _ I SPE _ C.

Pick up "A", how many places can you put this? 2.

Now we have _ I SPE A C.

Pick up "L", how many places can you put this? 1.

Now we have L I SPE A C.

We have exhausted the "letters" and that ends the process.

Thus total number of ways of arranging these "letters" is 5 x 4 x 3 x 2 x 1 = 5!

A question you may ask is why not add instead of multiply? or why multiply? This is because each event of placing down a "letter" is dependent. i.e. putting down say, "A", depends on what has already been placed down.

Think of multiplication as "and" and addition as "or". If you have studied set theory, then you can expand on this intuition greatly.

InteGrand · Apr 28, 2016

aanthnnyyy said:
Upcoming tutorial quiz coming up nxt week and probability still killing me from high school to uni could anyone please help me with these few Q's, textbook doesnt even display the answer. Please do note that im not just seeking solutions, i really want to know how to tackle these problems

1. How many ways can letters of the word SPECIAL be arranged such that the letters SPE appear together and in that order

2. A school has 500 pupils. A pupil who is absent has a probability of 0.2 of returning to school tomorrow. A pupil who is not absent today has a probability of 0.6 of being absent tomorrow. If 324 pupils are absent today, how many will be absent 3 days from now

3. Josh plays a game with three other friends. To win, Josh has to roll the die with a value higher than each his friends'. Josh's friends have values 7, 11 and 2 respectively. If Josh rolls a 20-sided die three times, what is the probability that he wins against all three friends

Here's some hints.

$\noindent 2) This can be modelled as a Markov process. Let $x_{1}(k)=\text{no. absent }k\text{ days from now}$ and $x_{2} (k) = \text{ no. at school }k\text{ days from now}$, for $k=0,1,2,\ldots$. Using the given transition probabilities, we can set up a system of coupled recurrence relations. It'll be of the form $\vec{x}\left(k+1\right) = A\vec{x}\left(k\right), \vec{x}(0) = \begin{bmatrix}324\\ 176\end{bmatrix}$, where $A$ will be the matrix of transition probabilities. We can then compute by hand recursively the values of the state vectors $\vec{x}\left(1\right),\vec{x}\left(2\right),$ and finally $\vec{x}\left(3\right)$. If we wanted a general form for $\vec{x}\left(k\right)$ as a function of $k$, we could do it by finding the eigenvalues of and diagonalising the matrix $A$, and then using the standard formula for the general solution of such a Markov process.$

$3) If this is interpreted as John needing to beat 7, then beat 11, then beat 2 (i.e. in that order) to win, the answer is easy to find (assuming of course the die rolls are independent and uniformly distributed etc.). If the question is interpreted as needing to beat 7,11 and 2, in any order, it is a bit harder. Here's a sketch for this harder method. It is based on finding the no. of ways he can win by considering cases.$

$\noindent It suffices to count the no. of ways where out of his three throws:$

$$\bullet$ All are 12 or higher$

$\bullet$ Two are 12 or higher and one is in $\{ 8,9,10,11 \}$

$\bullet$ Two are 12 or higher and one is in $\{ 3,4,5,6,7 \}$

$$\bullet$ One is 12 or higher and$

$\, \, \circ$ Two are in $\{8,9,10,11\}$

$$\, \, \circ$ One is in $\{8,9,10,11\}$ and one is in $\{3,4,5,6,7\}$.$

The above cases are mutually exclusive and exhaust all cases where John wins (I think...I may have left something out due to carelessness, but hopefully you get the idea). Counting the no. of ways to get each case is a mildly tedious but doable exercise.

Then to finally find the probability we want, divide the total no. of ways John can win by the total no. of ways he can roll. (There may also be a more elegant way of doing the problem than case-bashing.)

aanthnnyyy · Apr 28, 2016

InteGrand said:
Here's some hints.

$\noindent 2) This can be modelled as a Markov process. Let $x_{1}(k)=\text{no. absent }k\text{ days from now}$ and $x_{2} (k) = \text{ no. at school }k\text{ days from now}$, for $k=0,1,2,\ldots$. Using the given transition probabilities, we can set up a system of coupled recurrence relations. It'll be of the form $\vec{x}\left(k+1\right) = A\vec{x}\left(k\right), \vec{x}(0) = \begin{bmatrix}324\\ 176\end{bmatrix}$, where $A$ will be the matrix of transition probabilities. We can then compute by hand recursively the values of the state vectors $\vec{x}\left(1\right),\vec{x}\left(2\right),$ and finally $\vec{x}\left(3\right)$. If we wanted a general form for $\vec{x}\left(k\right)$ as a function of $k$, we could do it by finding the eigenvalues of and diagonalising the matrix $A$, and then using the standard formula for the general solution of such a Markov process.$

$3) If this is interpreted as John needing to beat 7, then beat 11, then beat 2 (i.e. in that order) to win, the answer is easy to find (assuming of course the die rolls are independent and uniformly distributed etc.). If the question is interpreted as needing to beat 7,11 and 2, in any order, it is a bit harder. Here's a sketch for this harder method. It is based on finding the no. of ways he can win by considering cases.$

$\noindent It suffices to count the no. of ways where out of his three throws:$

$$\bullet$ All are 12 or higher$

$\bullet$ Two are 12 or higher and one is in $\{ 8,9,10,11 \}$

$\bullet$ Two are 12 or higher and one is in $\{ 3,4,5,6,7 \}$

$$\bullet$ One is 12 or higher and$

$\, \, \circ$ Two are in $\{8,9,10,11\}$

$$\, \, \circ$ One is in $\{8,9,10,11\}$ and one is in $\{3,4,5,6,7\}$.$

The above cases are mutually exclusive and exhaust all cases where John wins (I think...I may have left something out due to carelessness, but hopefully you get the idea). Counting the no. of ways to get each case is a mildly tedious but doable exercise.

Then to finally find the probability we want, divide the total no. of ways John can win by the total no. of ways he can roll. (There may also be a more elegant way of doing the problem than case-bashing.)

Thanks IG ! Your feedback never fails

Edit: thanks paradoxica

1202 probabiility help please!!!! (1 Viewer)

aanthnnyyy

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parad0xica

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InteGrand

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