Help with a problem with probability sets

I can't wrap my mind around the answer:

Two events E and F ; the probability that neither is true is 0.6,

the probability that both are true is 0.2; find the probability that exactly one

of E or F is true.

So I started the problem with P(not E and not F)=.6, which means that P(E or F) = 1-.6=.4

and P(E and F) = .2

Exactly one is true is P(E\F) = P(E) - P(E and F), right? (or with P(F\E))

and by the laws of probability we have: P(E or F ) = P(E) + P(F) - P(E and F)

working the math down, I got to P(E) + P(F) = .6

At this point I'm stuck, I don't know where to go from here to figure out P(E\F). Any help would be greatly appreciated!

Re: Help with a problem with probability sets

Re: Help with a problem with probability sets

Thank you for pointing me in the right direction, and correcting my notation! I believe I have the right answer, I just wanted to run it by you (my apologies, I'm not familiar with Latex):

so P[E and F(c)] = P[E or F(c)] - P(E) - P(F)

& P[E(c) and F] = P[F or E(c)] - P(E) - P(F)

and added the two equations together. But P[E or F(c)] = P(F(c)) & P[F or E(c)] = P(E(c), so simplifying that I got:

P[E and F(c)] + P[E(c) and F] = P(F(c)) + P(E(c)) - 2(P(E)-P(F))

using the rule of compliments and P(E) + P(F) = .6 from above, this reduces to

P[E and F(c)] + P[E(c) and F] = 1- P(E) + 1 - P(F) -2(P(E)-P(F))

= 2 - 3[P(E)-P(F)] = 2 - 3(.6) = .2

Did I do that right or am I missing something?

Re: Help with a problem with probability sets

I cannot read you question.