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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) =
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) =

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[P(E \cup F)=P(E)+P(F)  P(E and F)\]Here P(E and F) is zero since the events are independent. So you have\[P(E \cup F)=P(E)+P(F) \rightarrow P(F)=P(E \cup F)P(E)=0.80.5=0.3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To three decimal places, 0.300

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to remember the first formula to get anywhere.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when i put that in as an answer it says it is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, I'm pretty sure that's the answer. Is this an online test or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no my probability class works off of a thing called wiley plus it gives you questions and i have four tries to get it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry...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
 5 years ago
Best ResponseYou've already chosen the best response.0No problem lol as long as you can help me some

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Technically,\[P(AB)=\frac{P(A and B)}{P(B)} \rightarrow = P(AB)P(B)=P(A and B)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, P(AB) = P(A) since it's independent of B, so P(E and F) = P(E)P(F) here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then\[P(E \cup F) = P(E)+P(F)+P(E)P(F)=P(E)+P(F)(1+P(E))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{P(E \cup F)P(E)}{1+P(E)}=P(F)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are these formulas coming through?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0somewhat I am trying to see the numbers going with it but im just confusing myself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P(F)=[P(E or F)P(E)]/[1P(E)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so P(F)=[0.8  0.5]/[10.5] = 0.3/0.5 = 0.600

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How many goes on your Wiley thing do you have left?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0good! i'm relieved...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because I made another mistake typing out, though on my notepad it's right...good luck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fan me if you like :) need the points.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay no problem take a look at some of the other problems i have up if you have the time
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