Distributions

ABSTRACT

A probability distribution defines how probability mass is allocated across a sample space $S$. It ensures that the likelihood of any specific outcome is between 0 and 1, and that the total likelihood of all possible outcomes sums exactly to 1.

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Probability Distributions

A distribution is a function $p:S\to [0,1]$ such that:

$$\sum _{a\in S}p(a)=1$$

Examples of Discrete Distributions

• Fair Coin: $S=\{H,T\}$, where $p(H)=1/2$ and $p(T)=1/2$.
• Fair Six-Sided Die: $S=\{1,2,3,4,5,6\}$, where $p(i)=1/6$ for all $i$.
• Standard Deck of Cards: $S=\{A♠,K♠,\dots \}$, where $p(\text{card})=1/52$.

Distribution of an Event

The probability of an event $E\subseteq S$ is the sum of the probabilities of the individual outcomes contained within that event:

$$p(E)=\sum _{a\in E}p(a)$$

• Certainty: $p(S)=1$
• Impossibility: $p(\varnothing )=0$

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The Law of Total Probability

The Law of Total Probability allows you to calculate the probability of an event $A$ by “weighting” it across different scenarios (a partition of the sample space).

Partitioning with Complements

For any two events $A$ and $B$, you can calculate $P(A)$ by looking at cases where $B$ occurs and where it does not ($\overline{B}$):

$$P(A)=P(A|B)P(B)+P(A|\overline{B})P(\overline{B})$$

General Form

If $\{B_1,B_2,\dots ,B_k\}$ is a partition of the sample space $S$ (meaning they are mutually exclusive and their union is $S$), then for any event $A$:

$$\boxed{P(A)=\sum _{i=1}^{k}P(A|B_i)P(B_i)}$$

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Example: Maximum of Two Dice

Question: What is the probability that the maximum of two dice rolls is greater than 4?

Let $A$ be the event that $\max ({\text{roll}}_1,{\text{roll}}_2)>4$. We can partition the space based on the first roll:

1. Event $B$: The first roll is $\{1,2,3,4\}$.
   • $P(B)=4/6=2/3$
   • $P(A|B)$: Given the first roll is $\le 4$, the second roll must be $5$ or $6$ for the max to be $>4$. Thus, $P(A|B)=2/6=1/3$.
2. Event $\overline{B}$: The first roll is $\{5,6\}$.
   • $P(\overline{B})=2/6=1/3$
   • $P(A|\overline{B})$: Since the first roll is already $>4$, the condition is satisfied regardless of the second roll. Thus, $P(A|\overline{B})=1$.

Calculation:

$$P(A)=\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)+(1)\left(\frac{1}{3}\right)=\frac{2}{9}+\frac{3}{9}=\boxed{\frac{5}{9}}$$

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Related Notes

• Conditional Probabilities: The Law of Total Probability is the denominator used in Bayes’ Theorem.
• Case Analysis: This is the practical application of the Law of Total Probability to simplify complex problems.
• Random Variables: Distributions of events form the basis for Probability Mass Functions (PMFs).