.. _review2:

Complement 2: Combinatorics and Probability Distribution
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.. note::

   This content was crafted by Ms. Colleen Pfaff.
   
Combinations
------------
.. math::
   C^n_r = \frac{n!}{(n-r)!r!}

Helpful Tip: If you are having trouble visualizing a probability problem draw it as a tree with each combination being a different branch.

Example Combinations
~~~~~~~~~~~~~~~~~~~~
Red cars numbered 2-10 and black cards 2-10 are placed in a bag. If 4 cards are selected at random, what is the probability that 2 are red and 2 are black?

Solution
~~~~~~~~
There are  :math:`C^9_2` ways to chose 2 red cards
There are  :math:`C^9_2` ways to chose 2 black cards

In order to calculate probability need to find out the total number of possible ways to choose 4 cards.

Number of ways to chose 4 cards :math:`C^{18}_4`

Probability = :math:`\frac{C^9_2C^9_2}{C^{18}_4} = \frac{1296}{3060}`

Discrete Probability
--------------------
.. math::
   <x> = \sum_i x_i P_i

.. math::
   <x^2> = \sum_i x_i^2P_i

Example Discrete Probability
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Calculate  :math:`<x>` and :math:`<x^2>` and :math:`\sigma^2` for a coin flip

Solution
~~~~~~~~
.. math::
   <x> = (1)(0.5) + (-1)(0.5) = 0
   
.. math::
   <x^2> = (1)^2(0.5) + (-1)^2(0.5) = 1

.. math::
   \sigma^2 = <x^2> - <x>^2 = 1 - 0 = 1

Continuous Probability
----------------------
.. math::
   <x> = \int x P(x)dx

and

.. math::
   \int P(x) = 1

Independent Variables
---------------------
.. math::
   <uv> = <u><v>

Example Independent Variables
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Are temperature and size of a system independent

Solution
~~~~~~~~
Yes temperature and size of system are independent!

Binomial Distribution
---------------------

Success has probability p and failure has probability 1-p. Success typically has a value of 1 and failure a value of -1.
These values for probability make sense because p + 1 - p = 1. 
Probability of k success(es) and n-k failures on n trials:

.. math::
   p^k(1-p)^{n-k}

If we have k success on n trials then there must be n-k failures because there are only two possible outcomes.
The average number of successful trials is the probability of success times the number of trials.

.. math::
   <k> = np

.. math::
   \sigma^2 = np(1-p)

Drunk's Walk
~~~~~~~~~~~~
A man leaves the bar and he is just as likely to step to the left as he is the right. What is his average distance after n steps and what is the variance?

Solution
~~~~~~~~
Let's see what the possibilities are after 2 steps: 
LL, RR, LR, RL and the probability of each of these results is

.. math::
   \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}

And this means that after 2 steps the average distance is 0
We can extrapolate this to N steps because this behavior won't change, therefore after N steps the average distance is 0.

.. math::
   <d^2> = (<s1> + <s2> + ... + <sN>)^2 = <s1^2> + <s2^2> + ... <sN^2> + <s1s2> + <s2s1> + \ldots

So the value of :math:`<s1^2> = 1`. Whereas the value of :math:`<s1s2>` or any other combination can take a couple of forms: (1)(1), (-1)(1), (1)(-1), (1)(1)

Since :math:`<s1s2>` or :math:`<s1sN>` is equally likely to be either 1 or -1 so like in the first step this will be 0

Therefore we have

.. math::
   <d^2> = (1 + 1 + 1 ... 1)

for every N step

.. math::
   <d^2> = N

.. math::
   \sigma_x = \sqrt{N}