nyuuzyou/wikis huggingface dataset:
Probability of All Players receiving Complete Suit at Bridge

== Theorem ==
The probability of all $4$ players in a game of Bridge being dealt a complete suit is $1$ in $2 \, 235 \, 197 \, 406 \, 895 \, 366 \, 368 \, 301 \, 560 \, 000$.
== Proof ==
== Historical Note ==
== Sources ==
* : $2,235,197,406,895,366,368,301,560,000$
* : $2,235,197,406,895,366,368,301,560,000$
Category:Bridge (Game)
Probability of All Players receiving Complete Suit at Bridge/Historical Note

== Historical Note on Probability of All Players receiving Complete Suit at Bridge ==
According to in his of $1986$, has pointed out "forcefully" that claims that all $4$ players have received a complete suit in a deal at Bridge has been made far more often than claims that only $2$ players have done so, even though the latter is far more likely than the former.
== Sources ==
* : $2,235,197,406,895,366,368,301,560,000$
* : $2,235,197,406,895,366,368,301,560,000$
Category:Historical Notes
Probability of All Players receiving Complete Suit at Bridge/Mistake

== Source Work ==
:
:The Dictionary
::$2,235,197,406,895,366,368,301,560,000$
== Mistake ==
:The probability that all $4$ players at Bridge will be dealt a complete suit.
== Correction ==
That should read:
:The reciprocal of the probability that all $4$ players at Bridge will be dealt a complete suit.
This has been corrected in of $1997$.
== Sources ==
* : $2,235,197,406,895,366,368,301,560,000$
Category:Mistakes/Curious and Interesting Numbers
Probability of Continuous Random Variable Lying in Singleton Set is Zero

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a continuous real variable on $\struct {\Omega, \Sigma, \Pr}$.
Then, for each $x \in \R$, we have: 
:$\map \Pr {X \le x} = \map \Pr {X < x}$
In particular: 
:$\map \Pr {X = x} = 0$
=== Corollary ===
== Proof ==
Let $F_X$ be the cumulative distribution function of $X$ so that: 
:$\map {F_X} x = \map \Pr {X \le x}$
for each $x \in \R$.
Let $P_X$ be the probability distribution of $X$. 
Since $X$ is a continuous real variable, we have: 
:$F_X$ is continuous. 
From Sequential Continuity is Equivalent to Continuity in the Reals, we have:
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$, we have $\map {F_X} {x_n}\to \map {F_X} x$.
For each $n \in \N$, let: 
:$\ds x_n = x - \frac 1 n$
We have that $x_n \to x$, so: 
:$\ds \map {F_X} x = \lim_{n \mathop \to \infty} \map {F_X} {x - \frac 1 n}$ 
We also have that $\sequence {x_n}_{n \mathop \in \N}$ is a increasing sequence. 
So, we have: 
:$\ds \hointl {-\infty} {x - \frac 1 n} \subseteq \hointl {-\infty} {x - \frac 1 {n + 1} }$
for each $n \in \N$. 
So:
:$\ds \sequence {\hointl {-\infty} {x - \frac 1 n} }_{n \mathop \in \N}$ is a increasing sequence of sets.
We can see that: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$
=== Lemma ===
So, we have: 
We then have: 
giving: 
:$\map \Pr {X = x} = 0$
== Also see ==
* Distribution Function of Finite Signed Borel Measure is Continuous iff Measure is Diffuse - of which this result is an instantiation with a similar proof
Category:Continuous Random Variables
Category:Probability of Continuous Random Variable Lying in Singleton Set is Zero
Probability of Continuous Random Variable Lying in Singleton Set is Zero/Corollary

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a continuous real variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $C$ be a countable subset of $\R$. 
Then: 
:$\map \Pr {X \in C} = 0$
== Proof ==
Since $C$ is countable, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ of distinct real numbers such that: 
:$C = \set {x_n : n \mathop \in \N}$
That is: 
:$\ds C = \bigcup_{n \mathop = 1}^\infty \set {x_n}$
where $\set {\set {x_1}, \set {x_2}, \ldots}$ is pairwise disjoint.
We then have: 
Category:Probability of Continuous Random Variable Lying in Singleton Set is Zero
Probability of Continuous Random Variable Lying in Singleton Set is Zero/Lemma

== Lemma ==
Let $x$ be a real number.
Then: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$
== Proof ==
We first show that: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} \subseteq \openint {-\infty} x$
Let: 
:$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$
Then, for some $k \in \N$ we have: 
:$\ds t \in \hointl {-\infty } {x - \frac 1 k}$
In particular, we have: 
:$t \le x - \dfrac 1 k$
Since: 
:$-\dfrac 1 k < 0$
we certainly have: 
:$t < x$
and hence: 
:$t \in \openint {-\infty} x$ 
So, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} \subseteq \openint {-\infty} x$
from the definition of subset.
We now show that: 
:$\ds \openint {-\infty} x \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$
Suppose that: 
:$t \in \openint {-\infty} x$
Then either: 
:$t \le x - 1$
or: 
:$x - 1 < t < x$
If: 
:$t \le x - 1$
then: 
:$\ds t \in \hointl {-\infty} {x - 1}$ 
so:
:$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$ 
Now suppose that: 
:$x - 1 < t < x$
Since:
:$x_n \to x$
and $\sequence {x_n}_{n \mathop \in \N}$ is increasing, there exists $N \in \N$ such that: 
:$\ds t \le x_N$
That is: 
:$\ds t \le x - \frac 1 N$
so:
:$\ds t \in \hointl {-\infty} {x - \frac 1 N}$
giving: 
:$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$
We therefore have: 
:$\ds \openint {-\infty} x \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n}$
So, from the definition of set equality, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$
Category:Probability of Continuous Random Variable Lying in Singleton Set is Zero
Probability of Empty Event is Zero

== Theorem ==
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
The probability measure $\Pr$ of $\EE$ has the following property:
:$\map \Pr \O = 0$
== Proof ==
From the conditions for $\Pr$ to be a probability measure, we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint.
From the definition of event space, we have:
:$\Omega \in \Sigma$
:$A \in \Sigma \implies \relcomp \Omega A \in \Sigma$
From Intersection with Empty Set:
:$\O \cap \Omega = \O$
Therefore $\O$ and $\Omega$ are pairwise disjoint.
From Union with Empty Set:
:$\O \cup \Omega = \Omega$
Therefore we have:
As $\map \Pr \Omega = 1$, it follows that $\map \Pr \O = 0$.
== Also see ==
* Elementary Properties of Probability Measure
== Sources ==
* : $1$: Events and probabilities: $1.3$: Probabilities: Example $10$
Category:Probability Theory
Probability of Event not Occurring

== Theorem ==
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\map \Pr A$ be the probability of event $A$ occurring.
Then:
:$\forall A \in \Sigma: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
That is, the probability of event $A$ not occurring is $1$ minus the probability of event $A$ occurring.
Another way of stating this is:
:$\map \Pr A + \map \Pr {\Omega \setminus A} = 1$
== Proof ==
From the conditions for $\Pr$ to be a probability measure, we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint.
Let $A \in \Sigma$ be an event.
Then $\paren {\Omega \setminus A} \in \Sigma$ by definition of Event Space: Axiom $(ES \ 2)$.
From Intersection with Relative Complement is Empty:
:$A \cap \paren {\Omega \setminus A} = \O$
From Union with Relative Complement:
:$A \cup \paren {\Omega \setminus A} = \Omega$
So:
:$\map \Pr A + \map \Pr {\Omega \setminus A} = 1$
from above, and so:
:$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
== Also see ==
* Elementary Properties of Probability Measure
== Sources ==
* : $1$: Events and probabilities: $1.3$: Probabilities: Example $10$
* : $1$: Events and probabilities: $1.4$: Probability spaces: $(14)$
Category:Probability Theory
Probability of Independent Events Not Happening

== Theorem ==
Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events in the event space of $\EE$.
Then the probability of none of $A_1$ to $A_m$ occurring is:
:$\ds \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
=== Corollary ===
== Proof ==
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events.
From Independent Events are Independent of Complement, we have that $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m \in \Sigma$ are also independent.
From the definition of occurrence, if $A$ does not happen then $\Omega \setminus A$ does happen.
So for none of $A_1, A_2, \ldots, A_m$ to happen, all of $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ must happen.
From Elementary Properties of Probability Measure:
:$\forall A \in \Omega: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
So the probability of none of $A_1$ to $A_m$ occurring is:
:$\ds \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
== Sources ==
* : $\S 1.7$: Independent Events: Exercise $24 \ \text{(i)}$
Category:Independent Events
Probability of Independent Events Not Happening/Corollary

== Theorem ==
Let $A$ be an event in an event space of an experiment $\EE$ whose probability space is $\struct {\Omega, \Sigma, \Pr}$.
Let $\map \Pr A = p$.
Suppose that the nature of $\EE$ is that its outcome is independent of previous trials of $\EE$.
Then the probability that $A$ does not occur during the course of $m$ trials of $\EE$ is $\paren {1 - p}^m$.
== Proof ==
This is an instance of Probability of Independent Events Not Happening with all of $A_1, A_2, \ldots, A_m$ being instances of $A$.
The result follows directly.
Category:Probability Theory
Probability of Limit of Sequence of Events/Decreasing

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\sequence {B_n}_{n \mathop \in \N}$ be a decreasing sequence of events.
Let $\ds B = \bigcap_{i \mathop \in \N} B_i$ be the limit of $\sequence {B_n}_{n \mathop \in \N}$.
Then:
:$\ds \map \Pr B = \lim_{n \mathop \to \infty} \map \Pr {B_n}$
== Proof ==
Set $A_i = \Omega \setminus B_i$ and then apply De Morgan's laws and the result for an increasing sequence of events.
== Sources ==
* : $\S 1.9$: Probability measures are continuous
Category:Probability Theory
Probability of Limit of Sequence of Events/Increasing

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence of events.
Let $\ds A = \bigcup_{i \mathop \in \N} A_i$ be the limit of $\sequence {A_n}_{n \mathop \in \N}$.
Then:
:$\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$
== Proof ==
Let $\ds B_i = A_i \setminus A_{i - 1}$ for $i \in \N: i > 0$.
Then:
:$A = A_0 \cup B_1 \cup B_2 \cup \cdots$
is the union of disjoint events in $\Sigma$.
By definition of probability measure:
But we have:
:$\map \Pr {B_i} = \map \Pr {A_i} - \map \Pr {A_{i - 1} }$ for $i \in \N: i > 0$.
So:
:$\ds \map \Pr A = \map \Pr {A_0} + \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \paren {\map \Pr {A_i} - \map \Pr {A_{i - 1} } }$
The last sum telescopes.
Hence the result:
:$\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$
== Sources ==
* : $\S 1.9$: Probability measures are continuous: Theorem $1 \ \text{C}$
Category:Probability Theory
Probability of Occurrence of At Least One Independent Event

== Theorem ==
Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events in the event space of $\EE$.
Then the probability of at least one of $A_1$ to $A_m$ occurring is:
:$\ds 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
=== Corollary ===
Let $A$ be an event in an event space of an experiment $\EE$ whose probability space is $\struct {\Omega, \Sigma, \Pr}$.
Let $\map \Pr A = p$.
Suppose that the nature of $\EE$ is that its outcome is independent of previous trials of $\EE$.
Then the probability that $A$ occurs at least once during the course of $m$ trials of $\EE$ is $1 - \paren {1 - p}^m$.
== Proof ==
Follows as a direct result of Probability of Independent Events Not Happening.
Let $B$ be the event "None of $A_1$ to $A_m$ happen".
From Probability of Independent Events Not Happening:
:$\ds \map \Pr B = \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
Then $\Omega \setminus B$ is the event "Not none of $A_1$ to $A_m$ happen", or "At least one of $A_1$ to $A_m$ happens".
From Elementary Properties of Probability Measure:
:$\forall A \in \Omega: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
Hence the probability that at least one of $A_1$ to $A_m$ happen is:
:$\ds 1 - \map \Pr B = 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
=== Proof of Corollary ===
It can immediately be seen that this is an instance of the main result with all of $A_1, A_2, \ldots, A_m$ being instances of $A$.
The result follows directly.
== Example ==
This is a classic result which contradicts the following equally classic fallacy:
:"There is a one in six chance of throwing a six with a single throw of a die.
:Therefore, there is a two in six chance of throwing a six on two throws of a die."
In fact this is an example of "occurrence of at least one independent event".
The probability of throwing at least one six on two throws of a die is in fact:
:$1 - \paren {1 - \dfrac 1 6}^2 = \dfrac {11} {36} < \dfrac 2 6$
Not a lot in it, but definitely significantly less.
See De Méré's Paradox for a real-world application of this result as it occurred in history.
Category:Probability Theory
Probability of Random Error under Normal Distribution occurring within Interval

== Theorem ==
Let $X$ be a random variable with a normal distribution.
Let:
:the expectation of $X$ be $0$
:the standard deviation of $X$ be $\sigma$.
Let $x \in \R$ be a real number.
Then the probability that a random error from $X$ lies in the interval $\openint {-x} x$ is given by:
:$p = \map \erf {\dfrac x {\sigma \sqrt 2} }$
where $\erf$ denotes the error function.
== Proof ==
== Sources ==
* 
Category:Error Function
Category:Normal Distribution
Category:Standard Deviation
Probability of Random Integer being Square-Free

== Theorem ==
Let $a$ be an integer chosen at random.
The probability that $a$ is square-free is given by:
:$\map \Pr {\neg \exists b \in \Z: b^2 \divides a} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$
where $\zeta$ denotes the zeta function.
The decimal expansion of $\dfrac 1 {\map \zeta 2}$ starts:
:$\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$
== Proof ==
Let $a$ be an integer chosen at random.
For $a$ to be square-free, it is necessary and sufficient that for all prime numbers $p$, it is not the case that $p^2$ is a divisor of $a$.
The probability that any particular integer is divisible by $p^2$ is $\dfrac 1 {p^2}$.
The probability that $a$ is not divisible by $p^2$ is therefore $1 - \dfrac 1 {p^2}$.
Whether or not $a$ is divisible by $p^2$ or divisible by $q^2$ for another prime number $q$ is independent of both $p$ and $q$.
Thus by the Product Rule for Probabilities, the probability that $a$ is not divisible by either $p^2$ or $q^2$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$.
This independence extends to all prime numbers.
That is, the probability that $a$ is not divisible by the square of any prime number is equal to the product of $1 - \dfrac 1 {p^2}$ over all prime numbers:
:$\map \Pr {\neg \exists b \in \Z: b^2 \divides a} = \ds \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
from which:
:$\ds \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
where $\map \zeta 2$ is the Riemann $\zeta$ (zeta) function evaluated at $2$.
The result follows from Riemann Zeta Function of 2.
== Also see ==
* Probability of Two Random Integers having no Common Divisor‎, which is the same probability as this
== Historical Note ==
== Sources ==
* : $0,60792 71018 \ldots$
* : $0 \cdotp 607 \, 927 \, 101 \ldots$
* : $0 \cdotp 60792 \, 7101 \ldots$
* 
Category:Riemann Zeta Function
Category:Square-Free Integers
Probability of Set Difference of Events

== Theorem ==
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$.
Let $\map \Pr A$ denote the probability of event $A$ occurring.
Then:
:$\map \Pr {A \setminus B} = \map \Pr A - \map \Pr {A \cap B}$
== Proof ==
From Set Difference and Intersection form Partition:
:$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
So, by the definition of probability measure:
:$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$
Hence the result.
== Sources ==
* : $1$: Events and probabilities: $1.4$: Probability spaces: Exercise $7$
Category:Probability Theory
Probability of Three Random Integers having no Common Divisor

== Theorem ==
Let $a, b$ and $c$ be integers chosen at random.
The probability that $a, b$ and $c$ have no common divisor:
:$\map \Pr {\map \perp {a, b, c} } = \dfrac 1 {\map \zeta 3}$
where $\zeta$ denotes the zeta function:
:$\map \zeta 3 = \dfrac 1 {1^3} + \dfrac 1 {2^3} + \dfrac 1 {3^3} + \dfrac 1 {4^3} + \cdots$
The decimal expansion of $\dfrac 1 {\map \zeta 3}$ starts:
:$\dfrac 1 {\map \zeta 3} = 0 \cdotp 83190 \, 73725 \, 80707 \ldots$
== Proof ==
Let $a$, $b$ and $c$ be three integers chosen at random.
For $a$, $b$ and $c$ to be coprime, it is necessary and sufficient that no prime number divides all of them.
The probability that any particular integer is divisible by a prime number $p$ is $\dfrac 1 p$.
Hence the probability that $a$, $b$ and $c$ are divisible by $p$ is $\dfrac 1 {p^3}$.
The probability that at least one of $a$, $b$ and $c$ is not divisible by $p$ is therefore $1 - \dfrac 1 {p^3}$.
Whether or not $a$ is divisible by $p$ or divisible by another prime number $q$ is independent of both $p$ and $q$.
Thus by the Product Rule for Probabilities, the probability that $a$, $b$ and $c$ are not all divisible by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^3} } \paren {1 - \dfrac 1 {q^3} }$.
This independence extends to all prime numbers.
That is, the probability that $a$, $b$ and $c$ are not all divisible by any prime number is equal to the product of $1 - \dfrac 1 {p^3}$ over all prime numbers:
:$\map \Pr {\map \perp {a, b, c} } = \ds \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^3} }$
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
from which:
:$\ds \dfrac 1 {\map \zeta 3} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^3} }$
where $\map \zeta 3$ is the Riemann $\zeta$ (zeta) function evaluated at $3$.
The result follows.
== Also see ==
* Probability of Two Random Integers having no Common Divisor
== Sources ==
* : $0 \cdotp 831 \, 907 \ldots$
* : $0 \cdotp 83190 \, 7 \ldots$
Category:Riemann Zeta Function
Category:Coprime Integers
Probability of Two Random Integers having no Common Divisor

== Theorem ==
Let $a$ and $b$ be integers chosen at random.
The probability that $a$ and $b$ are coprime is given by:
:$\map \Pr {a \perp b} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$
where $\zeta$ denotes the zeta function.
The decimal expansion of $\dfrac 1 {\map \zeta 2}$ starts:
:$\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$
== Proof ==
Let $a$ and $b$ be two integers chosen at random.
For $a$ and $b$ to be coprime, it is necessary and sufficient that no prime number divides both of them.
The probability that any particular integer is divisible by a prime number $p$ is $\dfrac 1 p$.
Hence the probability that both $a$ and $b$ are divisible by $p$ is $\dfrac 1 {p^2}$.
The probability that either $a$ or $b$ or both is not divisible by $p$ is therefore $1 - \dfrac 1 {p^2}$.
Whether or not $a$ is divisible by $p$ or divisible by another prime number $q$ is independent of both $p$ and $q$.
Thus by the Product Rule for Probabilities, the probability that $a$ and $b$ are not both divisible by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$.
This independence extends to all prime numbers.
That is, the probability that $a$ and $b$ are not both divisible by any prime number is equal to the product of $1 - \dfrac 1 {p^2}$ over all prime numbers:
:$\map \Pr {a \perp b} = \ds \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
from which:
:$\ds \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
where $\map \zeta 2$ is the Riemann $\zeta$ (zeta) function evaluated at $2$.
The result follows from Riemann Zeta Function of 2.
== Also see ==
* Probability of Random Integer being Square-Free‎, which is the same probability as this
== Historical Note ==
== Sources ==
* : $0,60792 71018 \ldots$
* : $0 \cdotp 607 \, 927 \, 101 \ldots$
* : $0 \cdotp 60792 \, 7101 \ldots$
* 
Category:Riemann Zeta Function
Category:Coprime Integers
Probability of Union of Disjoint Events is Sum of Individual Probabilities

== Theorem ==
Let $\EE$ be an experiment.
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability measure on $\EE$.
Then:
:$\forall A, B \in \Sigma: A \cap B = \O \implies \map \Pr {A \cup B} = \map \Pr A + \map \Pr B$
where $A \cap B$ denotes the union of $A$ and $B$.
== Proof ==
From the Kolmogorov Axioms:
:$\ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$
where $\set {A_1, A_2, \ldots}$ is a countable set of pairwise disjoint events of $\EE$.
This applies directly to $\map \Pr {A \cup B}$ where $A \cap B = \O$.
== Sources ==
* : $1$: Events and probabilities: $1.3$: Probabilities
* 
* 
* 
* 
Category:Addition Law of Probability
Category:Disjoint Events
Probability of no 2 People out of 53 Sharing the Same Birthday

== Theorem ==
Let there be $53$ people in a room.
The probability that no $2$ of them have the same birthday is approximately $\dfrac 1 {53}$.
== Proof ==
Let there be $n$ people in the room.
Let $\map p n$ be the probability that no two people in the room have the same birthday.
For simplicity, let us ignore leap years and assume there are $365$ days in the year.
Let the birthday of person $1$ be established.
The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$.
Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$.
Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$.
And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$.
Hence the total probability that none of the $n$ people share a birthday is given by:
:$\map p n = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$
Setting $n = 53$ and evaluating the above gives:
:$\map p {53} \approx 0.01887$
or:
:$\map p {53} \approx \dfrac 1 {53.01697}$
== Sources ==
* 
* : $53$
Category:Probability Theory
Category:53
Problem of Apollonius

== Classic Problem ==
Let three points, straight lines or circles, or any combination of $3$ such objects.
The problem is to construct a circle which passes through the points, or is tangent to the straight lines or circles.
=== Points ===
=== Straight Lines ===
=== Circles ===
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.6$: Apollonius (ca. B.C.)
* 
* 
Category:Problem of Apollonius
Category:Classic Problems
Category:Circles
Category:Plane Geometry