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Mathematics 18 Online
OpenStudy (anonymous):

If E and F are independent events, find P(F) if P(E) = 0.4 and = 0.7.

OpenStudy (amistre64):

your missing a part of the question i think

OpenStudy (anonymous):

i'm trying to find p(f) when p(e) = 0.4 and P (E u F) = 0.7

OpenStudy (zarkon):

use \[P(E\cup F)=P(E)+P(F)-P(E\cap F)\] along with independence

OpenStudy (anonymous):

i don't know how to do that

OpenStudy (zarkon):

one more hint.. \[.7=P(E)+P(F)-P(E)\cdot P(F)\]

OpenStudy (amistre64):

if p(e) and p(f) are independant; doesnt that mean that p(e n f) = 0?

OpenStudy (zarkon):

no

OpenStudy (amistre64):

venn diagram would be (E) (F) and they never the twain shall mix ?

OpenStudy (zarkon):

E and F are independent if and only if \[P(E\cap F)=P(E)\cdot P(F)\]

OpenStudy (zarkon):

you are mixing up independent and mutually exclusive

OpenStudy (anonymous):

so how would i set this up?

OpenStudy (zarkon):

look at what I have written above

OpenStudy (zarkon):

we have \[.7=P(E)+P(F)-P(E)\cdot P(F)\] and we know that \[P(E)=.4\] replace and solve for \[P(F)\]

OpenStudy (zarkon):

\[.7=P(E)+P(F)-P(E)\cdot P(F)\] \[.7=.4+P(F)-.4\cdot P(F)\] \[.3=P(F)-.4\cdot P(F)\] \[.3=P(F)(1-.4)\] \[.3=P(F)\cdot.6\] \[\frac{.3}{.6}=P(F)\] \[P(F)=.5\]

OpenStudy (anonymous):

i just did this all on my paper!!!!glad i saw i solved correctly!! thank you so much!!!!

OpenStudy (zarkon):

np

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