Complement 2: Combinatorics and Probability Distribution

Note

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Combinations

$C^n_r = \frac{n!}{(n-r)!r!}$

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 $C^9_2$ ways to chose 2 red cards There are $C^9_2$ 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 $C^{18}_4$

Probability = $\frac{C^9_2C^9_2}{C^{18}_4} = \frac{1296}{3060}$

Discrete Probability

$<x> = \sum_i x_i P_i$

$<x^2> = \sum_i x_i^2P_i$

Example Discrete Probability

Calculate $<x>$ and $<x^2>$ and $\sigma^2$ for a coin flip

Solution

$<x> = (1)(0.5) + (-1)(0.5) = 0$

$<x^2> = (1)^2(0.5) + (-1)^2(0.5) = 1$

$\sigma^2 = <x^2> - <x>^2 = 1 - 0 = 1$

Continuous Probability

$<x> = \int x P(x)dx$

and

$\int P(x) = 1$

Independent Variables

$<uv> = <u><v>$

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:

$p^k(1-p)^{n-k}$

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.

$<k> = np$

$\sigma^2 = np(1-p)$

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

$\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}$

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.

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