$ \renewcommand{\P}{\mathbf{P}} \def\given{\,\middle|\,} \newcommand{\setm}{\setminus} \newcommand{\Pr}[1]{\mathbf{P}\left(#1 \right)} \newcommand{\given}{\,\middle|\,} \newcommand{\qp}[1]{\left(#1\right)} \newcommand{\Var}[1]{\mathrm{Var}\left[#1\right\]} \newcommand{\E}[1]{\mathrm{E}\left[#1\right\]} $
Suppose $A$, $B$, and $C$ are three events in a random experiement with sample space $\Omega$. We will express a handful plain language events as sets in terms of the set operations union, intersection, complement and difference.
Let $\Omega^n$ be the discrete sample space for the outcomes of $n$ successive rounds of roulette, i.e., $$ \Omega^n = \{ 0, 00, 1, \ldots, 36\}^n. $$
For all simple events $x$ in $\Omega^n$, let $\P(x) = \frac{1}{36^n}$. As every subset $E$ of $\Omega^n$ can be written as the union of finitely many pairwise disjoint simple events $x$, we've defined $\mathbf{P}$ on all subsets of $\Omega$, thus defining a probability space $$ (\Omega^n, \text{power set of } \Omega^n, \mathbf{P}). $$
In $\Omega^2$, what's the probability that either the first round spins $12$ or the second round spins $18$?
Call this event $E$, where \begin{align} E &= \{ (a,b) \in \Omega^2 : a = 12 \text{ or } b = 18\}\\ &= \{ (12,b) \} \cup \{ (a,16) \}. \end{align}
Using the inclusion-exclusion formula, \begin{align} \P(E) &= \P(\{ (12,b) \} \cup \{ (a,16) \})\\ &= \P((12,b)) + \P((a,16)) - \P((12,16))\\ &= \frac{38}{38^n} + \frac{38}{38^n} - \frac{1}{38^n}\\ &= \frac{75}{1444}. \end{align}
What's the probability that the first spin is 12 or 18? Well, we're now in the space $\Omega^1$ with \begin{align} E &= \{ x \in \Omega^1 : x = 12 \text{ or } x = 18\}\\ &= \{12\} \cup \{18\}. \end{align}
We see that \begin{align} \P(E) &= \frac{1}{38} + \frac{1}{38}\\ & = \frac{1}{19}. \end{align}
Notice the events of spinning $12$ or spinning an $18$ are pairwise disjoint in $\Omega^1$. However, the events of spinning $12$ first and of spinning $18$ second in $\Omega^2$ are not pairwise disjoint.
Consider the sample space of obtaining either an $A$ or a $B$ in three courses, $\{A,B\}^3$. We can represent the elements in $\{A,B\}^3$ as leaves of the following binary tree:
courses = [[[],[]],[[],[]]],[[[],[]],[[],[]]]
ascii_art(BinaryTree(courses))
where each left branch denotes an $A$, and the (independent) probabilities for the first, second, and third left branches are $0.3$, $0.6$ and $0.8$ respectively. There are $2^3$ simple events in our sample space, but we're only after four:
Considering each simple event as a multistage experiment, we have
As each simple event (denote as $E_i$) is pairwise disjoint from the other three, so the probability of getting more $A$s than $B$s is the sum $$ \Pr{\bigcup_{\forall i} E_i} = \sum_{\forall i}\Pr{E_i} = 0.622. $$
Suppose that $\Omega$ is the sample space for the random experiment of choosing one worker at a factory. Let $A$ and $B$ be independent events in the space. Let $\P$ be a probaility measure on $\Omega$ such that $$ \Pr{A} = 0.65\\ \Pr{B} = 0.7\\ \Pr{A \cap B} = 0.455. $$
As the events $A$ and $B$ are independent, we have $\Pr{A \cap B} = \Pr{A}\Pr{B}$ (similarly for the complements of $A$ and $B$). Let's plot the known entries into a multiplacation table.
multiply | $1$ | $\Pr{A}$ | $\Pr{A^c}$ |
---|---|---|---|
$1$ | $1$ | $0.65$ | |
$\Pr{B}$ | $0.7$ | $0.455$ | $\quad\quad\quad$ |
$\Pr{B^c}$ | $\quad\quad\quad$ |
Normativity implies $1 = \Pr{A \cup A^c} = \Pr{A}+\Pr{A^c}$, and thus
multiply | $1$ | $\Pr{A}$ | $\Pr{A^c}$ |
---|---|---|---|
$1$ | $1$ | $0.65$ | $0.35$ |
$\Pr{B}$ | $0.7$ | $0.455$ | $\quad\quad\quad$ |
$\Pr{B^c}$ | $0.3$ | $\quad\quad\quad$ |
We find the remaining probabilities with the multiplication rule for independent events.
multiply | $1$ | $\Pr{A}\quad$ | $\Pr{A^c}\quad$ |
---|---|---|---|
$1$ | $1$ | $0.65$ | $0.35$ |
$\Pr{B}$ | $0.7$ | $0.455$ | $0.245$ |
$\Pr{B^c}$ | $0.3$ | $0.195$ | $0.105$ |
It follows that
Lastly, $$ \Pr{A \cup B} = \Pr{A} + \Pr{B} - \Pr{A \cap B} = 0.895. $$
Suppose an urn contains $6$ chips numbered from 1 to 6. Let $\Omega$ be the sample space for randomly drawing $3$ chips without replacement. Then $$|\Omega| = {6 \choose 3} = 20.$$
We'll find the probability of the event $E$ that largest chip drawn is the 4-chip.
Consider a multistage experiment.
We are finished; there are $1\times3$ ways to draw.
Whence $$\Pr{E} = \frac{|E|}{|\Omega|} = \frac{3}{20}.$$
Suppose I draw $4$ marbles without replacement from a bowl of $4$ red, $3$ white and $2$ blue.
Let the sample space $\Omega$ be the $9*8*7*6 = 3024$ permutations of length $4$ on a set of $9$ elements.
How many of these permutations contain at least $1$ marble of each color?
Consider the following multistage experiment.
There are $4*3*2*6 = 144$ permutations containing at least $1$ marble of each color.
Whence the probability of drawing at least $1$ marble of each color is $1/21$.
(Is this question nuanced? Am I missing a subtle point?---Maybe. I had come up with a different solution in terms of conditional probabilities, but, prefer the above approach on revision.)
With a probability of mowing his yard each Saturay, Samaniego exists for $5$ Saturdays.
To find the probality that he mows his lawn at least $4$ times, consider that
Each above combination is a simple event in our sample space, whence the desired probability is the sum $$ {5 \choose 5}\qp{\frac{2}{3}}^5+{5 \choose 4}\qp{\frac{2}{3}}^4\qp{\frac{1}{3}} \approx 0.461. $$
Each student in a population of $n = 10^3$ has a predictive dream with independent probability $p = 10^{-5}$ nightly. We'll determine the probability that at least one student has a predictive dream over $7$ nights.
In the sample space of $1$ night, the complement to the event that at least one student has a predictive dream is the probability that none do.
Whence, any night, the probability that at least $1$ student has a predictive dream is
\begin{align} 1 - {n \choose 0}p^0(1-p)^n &= 1 - \qp{1- \frac{1}{10^5}}^{10^3}\\ &= \frac{10^{5\cdot 10^3} - (10^3-1)^{10^3}}{10^{5\cdot 10^3}}, \end{align}call it $\alpha$.
Now, having established the probability for a night, we proceed to determine the probabilty for week's time, $N = 7$. Again, the probability that at least one student has a predictive dream in the sample space of a week is $1$ minus the probability that no one does (shame), i.e.,
\begin{align} 1 - {N \choose 0}\alpha^0(1-\alpha)^N &= 1 - \qp{1-{\alpha}}^{7}\\ &= \frac{10^{5\cdot 10^3\cdot 7} - (10^3-1)^{10^3\cdot 7}}{10^{5\cdot 10^3\cdot 7}}. \end{align}With later limit theorems, we can verify that this number is very close to
n = 10^3
# population
p = 10^(-5)
# incidence
a = 1 - e^(-n*p)
#1 minus the Poisson approx that no student dreams in a night
N(1 - binomial(7,0)*a^0*(1 - a)^7)
#1 minus prob that some student dreams in 7 nights
Suppose a committee consists of $8$ students, where each of $4$ disjoint classes have $2$ student representatives.
How many possible subcommittees of $4$ can be made from $8$?
In how many of the $70$ possible subcommittees are all $4$ classes represented?
In the case that the subcommittee is formed at random, what is the probability that $2$ of the classes would be completely excluded?
What is the probability that all $4$ suits are present in a random $5$ card poker hand?
Suppose that the kiddos, Jessica, Wolfgang and Frank, each have their own collection of cards from a random distribution of $20$ with replacement. Say that Jessica has $4$ cards, Wolfgang has $5$, and Frank has $6$. I'll assume that each kiddo doesn't have any repeats of one of their own cards.
To find the probability that, all together, they have $15$ unique cards, we'll represent each's collection as a function, then consider the probability that the images of any two functions overlap.
So, suppose that $$ \begin{align} j \colon &\{1,\ldots, 4\} \to \{1,\ldots,20\}\\ w \colon &\{5,\ldots, 9\} \to \{1,\ldots,20\}\\ f \colon &\{10,\ldots, 15\} \to \{1,\ldots,20\} \end{align} $$ are one-to-one functions.
Notice that there are $$ \begin{align} {20 \choose 4}& \text{ possible images of } j,\\ {20 \choose 5}& \text{ possible images of } w, \text{ and}\\ {20 \choose 6}& \text{ possible images of } f.\\ \end{align} $$
What's the probability that the function $g:\{1,\ldots,15\}\to\{1,\ldots,20\}$, defined $$ g(x) = \begin{cases} j(x), & \text{if } x \in \{1,\ldots, 4\} \\ w(x), & \text{if } x \in \{5,\ldots, 9\} \\ f(x), & \text{if } x \in \{10,\ldots, 15\} \end{cases} $$ is also one-to-one? That is, what's the probability the kids, all together, have $15$ unique cards?
(It can be shown $g$ is one-to-one if and only if the images of $j$, $w$, and $f$ are pairwise disjoint. We will focus our effort here once we define the sample space.)
First, notice that there are $$ {20 \choose 4}\cdot{20 \choose 5}\cdot{20 \choose 6} \approx 2.95115\times 10^{12} $$ different outcomes when we randomly determine the images of $j$, $w$, and $f$ (by the fundamental rule of counting). Let $\Omega$ be the sample space of all such outcomes.
Now, for how many simple events in $\Omega$ are the images of $j$, $w$, and $f$ pairwise disjoint? Well, as many as there are partitions of $20$ elements into sets of size $4$ (Jessica's), $5$ (Wolfgang's), $6$ (Frank's), and $5$ (nobody's).
We can use the multinomial coefficient. There are $$ {20 \choose 4,5,6,5} \approx 9.777 \times 10^9 $$ such partitions.
So, given that the images of $j$, $w$, and $f$ are randomly chosen, the probability that no two images overlap is the ratio $$ \frac{{20 \choose 4,5,6,5}}{{20 \choose 4}{20 \choose 5}{20 \choose 6}} = \frac{7007}{2086580} \approx 0.003358. $$
We have shown that the probability Jessica, Wolfgang and Frank have $15$ unique cards is about a third of a percent.