## anonymous 5 years ago If E and F are independent events, find P(F) if P(E) = 0.5 and P ( E u F) = 0.8. Round your answer to 3 decimal places. P(F) =

1. anonymous

$P(E \cup F)=P(E)+P(F) - P(E and F)$Here P(E and F) is zero since the events are independent. So you have$P(E \cup F)=P(E)+P(F) \rightarrow P(F)=P(E \cup F)-P(E)=0.8-0.5=0.3$

2. anonymous

To three decimal places, 0.300

3. anonymous

You have to remember the first formula to get anywhere.

4. anonymous

when i put that in as an answer it says it is wrong

5. anonymous

Hmm, I'm pretty sure that's the answer. Is this an online test or something?

6. anonymous

no my probability class works off of a thing called wiley plus it gives you questions and i have four tries to get it

7. anonymous

Hang on...

8. anonymous

kk

9. anonymous

Sorry...only had a few hours sleep and distracted...what we should have done is realized:$P(E and F)=P(E)P(F)$which isn't zero.

10. anonymous

No problem lol as long as you can help me some

11. anonymous

Technically,$P(A|B)=\frac{P(A and B)}{P(B)} \rightarrow = P(A|B)P(B)=P(A and B)$

12. anonymous

Now, P(A|B) = P(A) since it's independent of B, so P(E and F) = P(E)P(F) here.

13. anonymous

Then$P(E \cup F) = P(E)+P(F)+P(E)P(F)=P(E)+P(F)(1+P(E))$

14. anonymous

$\frac{P(E \cup F)-P(E)}{1+P(E)}=P(F)$

15. anonymous

Are these formulas coming through?

16. anonymous

somewhat I am trying to see the numbers going with it but im just confusing myself

17. anonymous

P(F)=[P(E or F)-P(E)]/[1-P(E)]

18. anonymous

so P(F)=[0.8 - 0.5]/[1-0.5] = 0.3/0.5 = 0.600

19. anonymous

How many goes on your Wiley thing do you have left?

20. anonymous

I had one more and that answer was correct so now i have to just figure out i think more problems and i will be good thanks so much for your help

21. anonymous

good! i'm relieved...

22. anonymous

because I made another mistake typing out, though on my notepad it's right...good luck

23. anonymous

fan me if you like :) need the points.

24. anonymous

okay no problem take a look at some of the other problems i have up if you have the time

25. anonymous

If I can, I will :)

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