mathmath333
  • mathmath333
Probablity question
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
Two 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
  • misty1212
i don't see the correct answer there
misty1212
  • misty1212
independent means \[P(A|B)=P(A)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathmath333
  • mathmath333
?
Zarkon
  • Zarkon
the answer is there
misty1212
  • misty1212
or \[P(A\cap B)=P(A)P(B)\] guess we need to look more carefully then huh?
misty1212
  • misty1212
ooh is that B just \[P(A'\cap B')=(1-P(A))(1-P(B))\]?
Zarkon
  • Zarkon
yes
misty1212
  • misty1212
sorry got confused with the proliferation of B's there
misty1212
  • misty1212
@Zarkon can explain why this one is the correct one
Zarkon
  • Zarkon
well I assume that is what they meant
misty1212
  • misty1212
process of elimination makes that what they meant
freckles
  • freckles
I think demorgan's theorem would be helpful
freckles
  • freckles
\[A' \text{ and } B'=(A \text{ or } B)'\]
freckles
  • freckles
\[P((A \text{ or } B)')=1-P(A \text{ or } B) \] then you can use that one thing for P(A or B)=P(A)+P(B)-P(A and B)
freckles
  • freckles
and misty said earlier P(A and B)=PA) P(B) since A and B are independent
Zarkon
  • Zarkon
P(A′B′) = [1 – P(A)] [1 – P(B)]=1-P(A)-P(B)+P(A)P(B) -1+P(A)+P(B)+P(A'B')=P(A)P(B) -1+(1-P(A'))+(1-P(B'))+P(A'B')=P(A)P(B) 1-(P(A')+P(B')-P(A'B')=P(A)P(B) 1-P(A'\(\cup\)B')=P(A)P(B) P(AB)=P(A)P(B)
misty1212
  • misty1212
i think you only need to know that \[1-P(A)=P(A')\]
anonymous
  • anonymous
like old home week today!

Looking for something else?

Not the answer you are looking for? Search for more explanations.