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10 questions from this quiz
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)
No, because 0.2≠0.5×0.40.2 \neq 0.5 \times 0.40.2=0.5×0.4
Pr(A∪B)=Pr(A)+Pr(B)\text{Pr}(A \cup B) = \text{Pr}(A) + \text{Pr}(B)Pr(A∪B)=Pr(A)+Pr(B)
When Pr(A)=0\text{Pr}(A) = 0Pr(A)=0 or Pr(B)=0\text{Pr}(B) = 0Pr(B)=0
No, these concepts are incompatible
000
12\frac{1}{2}21
0.18750.18750.1875
Yes, as shown in the dice example
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