## mathmath333 one year ago Probablity question

1. 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

2. misty1212

i don't see the correct answer there

3. misty1212

independent means $P(A|B)=P(A)$

4. mathmath333

?

5. Zarkon

6. misty1212

or $P(A\cap B)=P(A)P(B)$ guess we need to look more carefully then huh?

7. misty1212

ooh is that B just $P(A'\cap B')=(1-P(A))(1-P(B))$?

8. Zarkon

yes

9. misty1212

sorry got confused with the proliferation of B's there

10. misty1212

@Zarkon can explain why this one is the correct one

11. Zarkon

well I assume that is what they meant

12. misty1212

process of elimination makes that what they meant

13. freckles

I think demorgan's theorem would be helpful

14. freckles

$A' \text{ and } B'=(A \text{ or } B)'$

15. 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)

16. freckles

and misty said earlier P(A and B)=PA) P(B) since A and B are independent

17. 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)

18. misty1212

i think you only need to know that $1-P(A)=P(A')$

19. anonymous

like old home week today!