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) =

- anonymous

- schrodinger

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- 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\[P(E \cup F)=P(E)+P(F) \rightarrow P(F)=P(E \cup F)-P(E)=0.8-0.5=0.3\]

- anonymous

To three decimal places, 0.300

- anonymous

You have to remember the first formula to get anywhere.

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## More answers

- anonymous

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

- anonymous

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

- 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

- anonymous

Hang on...

- anonymous

kk

- 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.

- anonymous

No problem lol as long as you can help me some

- anonymous

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

- anonymous

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

- anonymous

Then\[P(E \cup F) = P(E)+P(F)+P(E)P(F)=P(E)+P(F)(1+P(E))\]

- anonymous

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

- anonymous

Are these formulas coming through?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

- anonymous

good! i'm relieved...

- anonymous

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

- anonymous

fan me if you like :) need the points.

- anonymous

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

- anonymous

If I can, I will :)

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