Probability

$P(cond)$ under x's distribution.

Probability(cond, Distributed(x, dist))exact for Equal/Less/LessEqual conditions on x (a chained range too); otherwise stays unevaluated.

Typed signatures

Type	Package	Applies
`(any, expression<Distributed>) -> real`	`statistics`	always

Domain: Statistics

Details

Equal(x, k): PDF $(k)$ for a discrete distribution, $0$ for a continuous one.
LessEqual(x, k): CDF $(k)$ . Less(x, k): $CDF(k) - PDF(k)$ for discrete (subtracting $P(X{=}k)$ ), just $CDF(k)$ for continuous (where a point has probability $0$ ). A constant on the LEFT (k \[LessEqual] x) is the complement of the opposite strict relation.
A chained range, a \[LessEqual] x \[LessEqual] b (and the Less/mixed forms), and And of two simple relations on x: composed from the primitives above by inclusion–exclusion.
Greater/GreaterEqual need no separate handling — compute-engine's own canonicalization rewrites x > k to Less(k, x) before this ever sees it.

Examples

Basic #

In

\mathrm{Probability}(x=2, \mathrm{Distributed}(x, \operatorname{PoissonDistribution}(3)))

Out

\frac{9}{2e^{3}}

In

\mathrm{Probability}(x=2, \mathrm{Distributed}(x, \operatorname{NormalDistribution}(0, 1)))

Out

0

$\approx 0.9160820579686966$ , cross-checked against wolframscript's N[CDF[PoissonDistribution[3],5]]

In

\mathrm{Probability}(x≤5, \mathrm{Distributed}(x, \operatorname{PoissonDistribution}(3)))

Out

\frac{92}{5e^{3}}

Scope #

$\approx 0.8662949896008326$ , cross-checked against wolframscript's N[Probability[1<=x<=5, x \[Distributed] PoissonDistribution[3]]]

In

\mathrm{Probability}(1≤x∧x≤5, \mathrm{Distributed}(x, \operatorname{PoissonDistribution}(3)))

Out

\frac{87}{5e^{3}}

compute-engine rewrites x > 0 to Less(0, x) before this evaluator sees it

In

N(\mathrm{Probability}(x\gt0, \mathrm{Distributed}(x, \operatorname{NormalDistribution}(0, 1))))

Out

0.5

Applications #

The chance of at least one hit, $1-(1-p)^n$ at $p=10^{-4}$

In

(()↦\mathrm{Simplify}(\operatorname{\_1}, n\gt0))(\mathrm{Probability}(1≤x, \mathrm{Distributed}(x, \operatorname{BinomialDistribution}(n, 1/10\,000))))

Out

1-(\frac{9\,999}{10\,000})^{n}

$P(X=0)=e^{-0.64}$ , as a number

In

N(\mathrm{Probability}(x=0, \mathrm{Distributed}(x, \operatorname{PoissonDistribution}(16/25))))

Out

0.527\,292\,424\,043\,048\,557\,244

The rate simplifies to 10, and $P(N=0)=e^{-10}$

In

\mathrm{Probability}(n=0, \mathrm{Distributed}(n, \operatorname{PoissonDistribution}(\frac{10\,000}{10^3})))

Out

e^{-10}

About

Examples	8
Checked against	Wolfram Language 3 agree

See also: Distributed, Expectation, CDF