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More Posts from Fade-22





I put you in the Apple car Ricky! I love you
Yay I'm going to drive to the library!

P(A)=Number of favorable outcomesTotal number of possible outcomesP(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}P(A)=Total number of possible outcomesNumber of favorable outcomes
P(A′)=1−P(A)P(A') = 1 - P(A)P(A′)=1−P(A)
P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)P(A∪B)=P(A)+P(B)
P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)P(A∪B)=P(A)+P(B)−P(A∩B)
P(A∣B)=P(A∩B)P(B)P(A | B) = \frac{P(A \cap B)}{P(B)}P(A∣B)=P(B)P(A∩B)
P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B)P(A∩B)=P(A)⋅P(B)
P(A∩B)=P(A)⋅P(B∣A)P(A \cap B) = P(A) \cdot P(B | A)P(A∩B)=P(A)⋅P(B∣A)
P(A∣B)=P(B∣A)⋅P(A)P(B)P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}P(A∣B)=P(B)P(B∣A)⋅P(A)
P(A)=∑i=1nP(A∣Bi)⋅P(Bi)P(A) = \sum_{i=1}^{n} P(A | B_i) \cdot P(B_i)P(A)=∑i=1nP(A∣Bi)⋅P(Bi)
p(x)p(x)p(x): E(X)=∑xx⋅p(x)E(X) = \sum_{x} x \cdot p(x)E(X)=∑xx⋅p(x)
Var(X)=E(X2)−[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2Var(X)=E(X2)−[E(X)]2
σX=Var(X)\sigma_X = \sqrt{\text{Var}(X)}σX=Var(X)
P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}P(X=k)=(kn)pk(1−p)n−k where (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}(kn)=k!(n−k)!n!
P(X=k)=λke−λk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}P(X=k)=k!λke−λ
f(x)=1σ2πe−(x−μ)22σ2f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}f(x)=σ2π1e−2σ2(x−μ)2
