Test Cheat Sheet

Baye’s theorem

$P(A_i|B)=\frac{P(A_i\cap B)}{P(B)}=\frac{P(B|A_i)P(A_i)}{{\sum }_{j=1}^{k}P(B|A_j)P(A_j)}$ Where $P(A_i)$ is the prior probability of the event $A_i$, which means this is the unconditional probability before considering additional information.

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Permutations

An ordered subset of distinct choices is called a permutation, and $P_{n,k}=\text{num. of permutations of size}k\text{that can be formed from}n\text{objects}$

General formula: $n$ factorial over $(n-k)$ factorial

$$P_{n,k}=\frac{n!}{(n-k)!}$$

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Combinations

An unordered subset of choices is called a combination, and $C_{n,k}=\text{num. of combinations of size}k\text{that can be formed from}n\text{objects}$

Often written as $C_{n,k}$ can be written $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, which can be read as ”$n$ choose $k$”

General Formula:

$$C_{n,k}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{(n-k)!k!}$$

It’s also good to know that $n$ choose $k$ is equivalent to $n$ choose $n-k$, because the bottom term remains the same.

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Independence Checking

Method 1

Check $P(A|B)=P(A)$ or $P(B|A)=P(B)$

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Method 2

Check $P(A\cap B)=P(A)P(B)$

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Distinguishable Objects and Distinguishable Boxes

The number of ways of distributing $n$ distinguishable objects into $r$ distinguishable boxes so that $n_i$ objects are placed into box $i$, for $i=1,2,\dots ,r$ is:

$$\left(\genfrac{}{}{0ex}{}{n}{n_1,n_2,\dots ,n_r}\right)=\frac{n!}{n_1!n_2!\dots n_r!}$$ Link to original

Indistinguishable Objects and Distinguishable Boxes

Every box gets an object

Suppose we want to distribute $n$ indistinguishable objects into $k$ boxes so that every box gets at least one object

If $x_i$ is the number of objects going into box $i$, we are looking for the number of solutions to:

$$x_1+x_2+\dots +x_k=n,\text{where }{x}_i>0$$

If we express this as bars and stars, you want to place $k-1$ bars in $n-1$ spaces, so this can be expressed as:

$$\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$$

This is just regular combinations.

Not every box gets an object

Suppose we want to distribute $n$ indistinguishable objects into $k$ boxes so that every box may or may not get an object

If $x_i$ is the number of objects going into box $i$, we are looking for the number of solutions as the sum of $k$ non-negative integers:

$$x_1+x_2+\dots +x_k=n,\text{where }{x}_i\ge 0$$

The number of ways we can do this can be expressed as:

$$\left(\genfrac{}{}{0ex}{}{n+k-1}{k-1}\right)$$Link to original

Probability Laws

1. If $A\subseteq B$, then $P(A)\le P(B)$
2. $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
   1. “General Addition Rule” (Probability of the union of events)
   2. If A and B are mutually exclusive, the rule is $P(A\cup B)=P(A)+P(B)$
3. $P(A\cup B)\le P(A)+P(B)$
4. $P(A\cup B\cup C)=P(A)+P(A^{\complement }\cap B)+P(A^{\complement }\cap B^{\complement }\cap C)$

Probability of a complement: $P(A^{\complement })=1-P(A)$

Probability of the null event: $P(\varnothing )=0$

Partitioning an event: $P(A)=P(A\cap B)+P(A\cap B^{\complement })$

Probability of any two disjoint events: For $A\text{and}B,P(A\cap B)=0$

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