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) =
\[P(E \cup F)=P(E)+P(F) - P(E and F)\]Here P(E and F) is zero since the events are independent. So you have\[P(E \cup F)=P(E)+P(F) \rightarrow P(F)=P(E \cup F)-P(E)=0.8-0.5=0.3\]
To three decimal places, 0.300
You have to remember the first formula to get anywhere.
when i put that in as an answer it says it is wrong
Hmm, I'm pretty sure that's the answer. Is this an online test or something?
no my probability class works off of a thing called wiley plus it gives you questions and i have four tries to get it
Hang on...
kk
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.
No problem lol as long as you can help me some
Technically,\[P(A|B)=\frac{P(A and B)}{P(B)} \rightarrow = P(A|B)P(B)=P(A and B)\]
Now, P(A|B) = P(A) since it's independent of B, so P(E and F) = P(E)P(F) here.
Then\[P(E \cup F) = P(E)+P(F)+P(E)P(F)=P(E)+P(F)(1+P(E))\]
\[\frac{P(E \cup F)-P(E)}{1+P(E)}=P(F)\]
Are these formulas coming through?
somewhat I am trying to see the numbers going with it but im just confusing myself
P(F)=[P(E or F)-P(E)]/[1-P(E)]
so P(F)=[0.8 - 0.5]/[1-0.5] = 0.3/0.5 = 0.600
How many goes on your Wiley thing do you have left?
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
good! i'm relieved...
because I made another mistake typing out, though on my notepad it's right...good luck
fan me if you like :) need the points.
okay no problem take a look at some of the other problems i have up if you have the time
If I can, I will :)
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