Tag Info

New answers tagged

3

Fixed in 10.0.2 Probability[a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 < 1, {a, b, c, d, e, f, g} \[Distributed] UniformDistribution[{{0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}}]]


2

pmd = ParameterMixtureDistribution[BinomialDistribution[n, p], {n, p} \[Distributed] ProductDistribution[PoissonDistribution[λ], BetaDistribution[α, β]]] Mean[pmd] $ \frac{\alpha \lambda }{\alpha +\beta } $ Variance[pmd] $\frac{\alpha ^3 \lambda +2 \alpha ^2 \beta \lambda +\alpha ^2 \lambda +\alpha \beta ^2 \lambda +\alpha \beta ...


0

Question If $X \sim N(0,1)$ and $Y \sim N(0,1)$ are independent, how come: $$ P(X<Y) = \frac12 \quad \text{while} \quad P(\frac{X}{Y}<1) = \frac34$$ Answer Nothing to do with Mma. The answer is that the rules of standard algebra do NOT apply to the algebra of random variables, so you cannot simply divide both parts of $P(X<Y)$ by $Y$ ... ...


2

As this is simple enough, you could use definitions : dist = MultinormalDistribution[{0, 0, 0}, IdentityMatrix[3]] ; prob[a_] = Integrate[PDF[dist, {z1, z2, z3}], {z2, -Infinity, Infinity}, {z3, -Infinity, Infinity}, {z1, a Sqrt[z2^2 + z3^2], Infinity}] (* ConditionalExpression[1/4 (2 - ...


3

First, let's find a in functional form: Probability[Abs[x] < a, x \[Distributed] NormalDistribution[]] which shows that f[a_] := Erf[a/Sqrt[2]] Now you want to solve for the a that has some probability, say 0.1. Then Solve[f[a] == 0.1, a] {{a -> 0.125661}} A little more generally: Solve[f[a] == p, a] {{a -> Sqrt[2] InverseErf[p]}} ...


0

You probably mean this question for mathematics rather than Mathematica, but you can use Mathematica to brute force the probabilities: d2 = Flatten@Outer[Plus, Range@6, Range@6]; d3 = Flatten@Outer[Plus, Range@6, Range@6, Range@6]; d4 = Flatten@Outer[Plus, Range@6, Range@6, Range@6, Range@6]; Chance of winning with 3 dice versus 2 dice: Total@#/Length@# ...


2

You need higher precision to finish these calculations badset = N[{alpha -> 10^(-14), beta -> (114/100)*10^(26), p -> 0, w -> 0}, 30]; probs /. badset {1.0, 4.*10^-40, 8.*10^-66, 1.3*10^-91, 1.8*10^-117, 1.2*10^-143} And Sum[%[[i]], {i, 1, 6}] 1.0



Top 50 recent answers are included