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10 cards from this deck
One event's occurrence doesn't affect other's probability
Pr(A∣B)=Pr(A)\text{Pr}(A | B) = \text{Pr}(A)Pr(A∣B)=Pr(A)
Pr(A∩B)=Pr(A)×Pr(B)\text{Pr}(A \cap B) = \text{Pr}(A) \times \text{Pr}(B)Pr(A∩B)=Pr(A)×Pr(B)
Multiply: Pr(A)×Pr(B)\text{Pr}(A) \times \text{Pr}(B)Pr(A)×Pr(B)
Check if Pr(A∩B)=Pr(A)×Pr(B)\text{Pr}(A \cap B) = \text{Pr}(A) \times \text{Pr}(B)Pr(A∩B)=Pr(A)×Pr(B)
Pairwise independence
Independent: no effect; Mutually exclusive: can't both occur
No, unless at least one has probability zero
Physical independent → mathematical independent, not reverse
Pr(A)+Pr(B)−Pr(A)×Pr(B)\text{Pr}(A) + \text{Pr}(B) - \text{Pr}(A) \times \text{Pr}(B)Pr(A)+Pr(B)−Pr(A)×Pr(B)
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