Finding $P(B)$ if $A,B$ are independent and given values for $P(A),P(A\cup B)$ I am trying to figure out the math to get to an answer that's given to me right away, which is $1/3$. The question is asking what the probability of $B$ is with $P(A)=0.4$ and $P(A∪B)=0.6$, with $A,B$ being independent events. 
I can't seem to come to one-third? Can someone explain it to me like I'm a three year old, since I have been on this for about a few hours? 
Thanks, and much appreciated! Below is the setup I keep using:
$$P(A∪B)=P(A)+P(B)-P(A∩B)$$
which, when substituting in the given values and letting $P(B) = x$, yields
$$0.6 = 0.4 + x - P(A \cap B)$$
But I don't know what $P(A \cap B)$ is, I guess $0.4x$?
 A: We know that, if $A,B$ are independent, then
$$P(A \cap B) = P(A)P(B)$$
We also know, through inclusion-exclusion,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Combining the two, if we use inclusion-exclusion for $A,B$ independent,
$$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$
In your problem, you are given $P(A \cup B)$ and $P(A)$. We note by factoring out the $P(B)$ in the previous expression,
$$P(A \cup B) = P(A) + P(B)(1 - P(A))$$
You should be able to take it from here, now it's just algebra.

Edit: (the remainder of the solution to address confusion that came up in later comments on other answers)
So, we solve for $P(B)$ by subtracting $P(A)$ from both sides, then dividing by $1 - P(A)$, thus getting
$$P(B) = \frac{P(A\cup B) - P(A)}{1 - P(A)}$$
We plug in the given values - $P(A) = 0.4, P(A \cup B) = 0.6$:
$$P(B) = \frac{0.6 - 0.4}{1 - 0.4} = \frac{0.2}{0.6} = \frac 1 3$$
A: Yes, if $A$ and $B$ are independent, then $\mathsf P(A\cap B)=\mathsf P(A)\,\mathsf P(B)$ .
Hence the formula you seek is $\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A)~\mathsf P(B)$ rearranged to: $$\mathsf P(B)=\dfrac{\mathsf P(A\cup B)-\mathsf P(A)}{1-\mathsf P(A)}$$
A: By definition if $A, B$ are independent events then $P(A \cap B) = P(A) P(B)$. Substituting in the inclusion-exclusion formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ you wrote gives
$$P(A \cup B) = P(A) + P(B) - P(A) P(B) .$$
Now, you're given $P(A \cup B) = \frac{3}{5}$ and $P(A) = \frac{2}{5}$ substituting in the above equation gives an equation in $P(B)$ alone.
A: If P(A) and P(B) are independent then P(A)P(B)=P(AnB)
P(B\A)=P(AUB)-P(A)=0.2
We know P(A)P(B)=0.4(0.2+P(AnB))=P(AnB)
Hence P(AnB)=2/15 and so P(B)=P(B\A)+P(AnB)=1/3
A: Okay, I just found out what is wrong with what I am trying to do. Remember that -.4x? Combine that with x. Then, subtract the .4 from .6, and divide. So something like this, if I do the proper bold-ing and italics:
.6 = .4 + x - .4x
.6 = .4 + .6x
.2 = .6x
.33 = x
There, that should make more sense. Thank you guys for the help! That was actually really helpful, now I know something more from all of your help, so have a great day! 
