Basic Probability

Event, Sample Space, and Probability

In probability theory, an experiment is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space, $S$. We define an event to be a set of outcomes of an experiment (also known as a subset of $S$) to which a probability (numerical value) is assigned.

The probability of the occurrence of an event, $A$, is:

$\huge P(A) =\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Here are the first two fundamental rules of probability:

1. Any probability, $P(A)$, is a number between $0$ and $1$ (i.e., $0\leq P(A) \leq1$).

2. The probability of the sample space, $S$, is $1$ (i.e., $P(S) = 1$).

So how do we bridge the gap between the value of $P(A)$ and the sample space? Quite simply, since we know that $P(A)$ is the probability that event $A$ will occur, then we define $P(A')$ (also written as $P(A^c)$) to be the probability that event will not occur (the complement of $P(A)$). If our sample space is composed of the probabilities of 's occurrence and non-occurrence, we can then say $P(A) + P(A')=1$, or the sum of all possible outcomes of $A$ in the sample space is equal to $1$. This is the third fundamental rule of probability: $\newline P(A^c)=1 - P(A)$.

Question 1 Find the probability of getting an odd number when rolling a $6$-sided fair die.

Given the above question, we can extract the following:

• Experiment: rolling a $6$-sided die.

• Sample space ($S$): $S=\{1, 2, 3, 4, 5, 6\}$.

• Event ($A$): that the number rolled is odd (i.e., $A=\{1, 3, 5\}$).

If we refer back to the basic formula for the probability of the occurrence of an event, we can say: $\large P(A) =\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{\vert{A}\vert}{\vert{S}\vert} =\frac{3}{6}=\frac{1}{2}$

Compound Events, Mutually Exclusive Events, and Collectively Exhaustive Events

Let's consider $2$ events: $A$ and $B$. A compound event is a combination of $2$ or more simple events. If $A$ and $B$ are simple events, then $A\cup B$ denotes the occurrence of either $A$ or $B$. Similarly, $A\cap B$ denotes the occurrence of $A$ and $B$ together.

$A$ and $B$ are said to be mutually exclusive or disjoint if they have no events in common (i.e., $A \cap B = \emptyset$ and $P(A \cap B) = 0$). The probability of any of $2$ or more events occurring is the union ($\cup$) of events. Because disjoint probabilities have no common events, the probability of the union of disjoint events is the sum of the events' individual probabilities. $A$ and $B$ are said to be collectively exhaustive if their union covers all events in the sample space (i.e., $A \cup B = S$ and $P(A \cup B) = 1$). This brings us to our next fundamental rule of probability: if $2$ events, $A$ and $B$, are disjoint, then the probability of either event is the sum of the probabilities of the $2$ events (i.e., $P(A \ or B) = P(A) + P(B)$).

If the outcome of the first event ($A$) has no impact on the second event ($B$), then they are considered to be independent (e.g., tossing a fair coin). This brings us to the next fundamental rule of probability: the multiplication rule. It states that if two events, $A$ and $B$, are independent, then the probability of both events is the product of the probabilities for each event (i.e., $P(A \ and\ B) = P(A) * P(B)$). The chance of all events occurring in a sequence of events is called the intersection ($\cap$) of those events.

Question 2 Find the probability of getting $1$head and $1$ tail when $2$ fair coins are tossed.

Given the above question, we can extract the following:

• Experiment: tossing $2$ coins.

• Sample space ($S$): The possible outcomes for the toss of $1$ coin are $\{H, T\}$, where $H = heads$ and $T = tails$. As our experiment tosses $2$ coins, we have to consider all possible toss outcomes by finding the Cartesian Product of the possible outcomes for each coin: $\newline S = \{\{ H, T\} *\{H, T\} = \{(H,H),(H,T),(T,H),(T,T)\}$.

• Event ($A \cap B$): that the outcome of $1$ toss will be $H$, and the outcome of the other toss will be $T$(i.e., $A=\{(H, T), (T, H)\}$.

Connecting this information back to our basic formula for $P(A)$, we can say:

$\large P(A) =\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{\vert{A}\vert}{\vert{S}\vert} =\frac{2}{4}=\frac{1}{2}$

Question 3 Let $A$ and $B$ be two events such that $P(A)=\frac{2}{5}$ and $P(B)=\frac{4}{5}$. If the probability of the occurrence of either $A$ or $B$ is $\frac{3}{5}$, find the probability of the occurrence of both $A$ and $B$ together (i.e., $A \cap B$).

We can use our fundamental rules of probability to solve this problem:

$\vert A \cup B\vert = \vert A \vert + \vert B \vert - \vert A \cap B\vert \newline \Rightarrow P(A \cup B) = P(A)+P(B)-P(A \cap B)\newline \Rightarrow P(A \cap B) = P(A)+P(B)-P(A \cup B)\newline \Rightarrow P(A \cup B) = \frac{2}{5}+\frac{4}{5} -\frac{3}{5}=\frac{3}{5}$

Task

There are urns labeled $X$, $Y$, and $Z$.

• Urn $X$ contains $4$ red balls and $3$ black balls.

• Urn $Y$ contains $5$ red balls and $4$ black balls.

• Urn $Z$ contains $4$ red balls and $4$ black balls.

One ball is drawn from each of the $3$ urns. What is the probability that, of the $3$ balls drawn, $2$ are red and $1$ is black?

Solution

Urn X has a 4/7 probability of giving a red ball Urn Y has a 5/9 probability of giving a red ball Urn Z has a 1/2 probability of giving a red ball

Urn X has a 3/7 probability of giving a black ball Urn Y has a 4/9 probability of giving a black ball Urn Z has a 1/2 probability of giving a black ball

P(2 red, 1 black) = P(Red Red Black) + P(Red Black Red) + P(Black Red Red) = (4/7)(5/9)(1/2) + (4/7)(4/9)(1/2) + (3/7)(5/9)(1/2) = 20/126 + 16/126 + 15/126 = 51/126 = 17/42