# Math Help - Help with a problem with probability sets

1. ## 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; fi nd 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!

2. ## Re: Help with a problem with probability sets

Originally Posted by puggles
Two events E and F ; the probability that neither is true is 0.6, the probability that both are true is 0.2; fi nd 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!
There are several problems with this posting.
First, we do not use the word true when referring to events.
We say that an event occurs or not.
Notation: $\mathcal{P}(E^c)$ is the probability that $E$ does not occur.

So neither occurs is $\mathcal{P}(E^c\cap F^c)=0.6$
Thus $\mathcal{P}(E\cup F)=0.4$, at least one occurs.
More given: $\mathcal{P}(E\cap F)=0.2$.

Now here is you problem:
Find $\mathcal{P}(E\cap F^c)+\mathcal{P}(E^c\cap F)=~?$
That is exactly one occurs.

3. ## 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?

4. ## Re: Help with a problem with probability sets

$\mathcal{P}(E^c} )=1-\mathcal{P}(E})$