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mathmath333
 one year ago
Probablity question
mathmath333
 one year ago
Probablity question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0Two events A and B will be independent, if (A) A and B are mutually exclusive (B) (B) P(A′B′) = [1 – P(A)] [1 – P(B)] (C) P(A) = P(B) (D) P(A) + P(B) = 1

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1i don't see the correct answer there

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1independent means \[P(AB)=P(A)\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1or \[P(A\cap B)=P(A)P(B)\] guess we need to look more carefully then huh?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1ooh is that B just \[P(A'\cap B')=(1P(A))(1P(B))\]?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1sorry got confused with the proliferation of B's there

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1@Zarkon can explain why this one is the correct one

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1well I assume that is what they meant

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1process of elimination makes that what they meant

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think demorgan's theorem would be helpful

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[A' \text{ and } B'=(A \text{ or } B)'\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[P((A \text{ or } B)')=1P(A \text{ or } B) \] then you can use that one thing for P(A or B)=P(A)+P(B)P(A and B)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and misty said earlier P(A and B)=PA) P(B) since A and B are independent

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1P(A′B′) = [1 – P(A)] [1 – P(B)]=1P(A)P(B)+P(A)P(B) 1+P(A)+P(B)+P(A'B')=P(A)P(B) 1+(1P(A'))+(1P(B'))+P(A'B')=P(A)P(B) 1(P(A')+P(B')P(A'B')=P(A)P(B) 1P(A'\(\cup\)B')=P(A)P(B) P(AB)=P(A)P(B)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1i think you only need to know that \[1P(A)=P(A')\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like old home week today!
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