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Let E and F be two independent events with $P(E) = 0.4$ and $P(E \cup F) = 0.64$ What is $P( F)$?

So I've tried to figure this one out but I just can't.

I got as far as $P(E \cup F) = P(E) + P( F) - P(E)P( F)$

                       $0.64 = 0.4 + P( F) - 0.4P( F)$

but I don't really know how I can go anywhere from there.

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    you are almost there, assume $P(F) = x$ and re-arrange for $x$ – 2017-01-26
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    Factor the term you are looking for to get $0.64 = 0.4 + P(F)(1-0.4)$. – 2017-01-26

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You're literally almost there.

Treat P(F) as a variable, so you get $0.64-0.4=(1-0.4)P(F)$, so $0.24=0.6P(F).$ So P(F) is just $\frac{0.24}{0.6}=\boxed{0.4}.$

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    OHHH i get it thanks – 2017-01-26