What's the probability of an event happening exactly once after two independent trials? The problem:

If event $A$ happens with probability $p$, what is the probability $P[K=1]$ that, after two independent trials, $A$ happened exactly once?

Is the answer:


*

*$P[K=1] = p(1-p)$, since we need $A$ to happen and then we need $A$ not to happen, or

*$P[K=1] = 2p(1-p) = p(1-p) + (1-p)p$, since there are two cases $A\overline{A}$ and $\overline{A}A$ for which $A$ happen  exactly once?

 A: Ok, well there are 4 possible outcomes 
$$ {\rm success/success}, \ {\rm success/failure}, \ {\rm failure/success},\ {\rm failure/failure},  $$ 
which occur with probabilities 
$$p^2, \ \ \ p(1-p), \ \ \ (1-p)p \ \ {\rm and} \ \ (1-p)^2$$
respectively, assuming the trials are independent. There is exactly one success in the cases ${\rm success/failure}$ and ${\rm failure/success}$. Since order doesn't matter, 
$$ P({\rm exactly \ one \ success}) = P({\rm success/failure \ OR \ failure/success}) $$
In general, for two events $A,B$ 
$$P(A \ \ {\rm OR} \ \ B) = P(A) + P(B) - P(A \ \ {\rm AND} \ \ B) $$
In your example
$$P(A) = P({\rm success/failure}) = p(1-p)$$
$$P(B) = P({\rm failure/success}) = p(1-p)$$   
since ${\rm success/failure}$ (event A) and ${\rm failure/success}$ (event B) are clearly disjoint events
(it's impossible for events A and B to occur together), their intersection is empty:
$$P(A \ \ {\rm AND} \ \ B) = 0 $$  
thus 
$P(A \ \ {\rm OR} \ \ B) = P(A) + P(B) - P(A \ \ {\rm AND} \ \ B)$
$P(A \ \ {\rm OR} \ \ B) = p(1-p) + (1-p)p - 0$
$P(A \ \ {\rm OR} \ \ B) = 2p(1-p) $
which was your second option. 
A: The first option is wrong because the question does not state which trial $A$ occurs.  If you say what is the probability that $A$ occurs on the first trial but not on the second, then the first option is right. Now if you say what is the probability that if $A$ occurs on the first trial it will not repeat.  That probability would be $1-p$.  This is really a very elementary problem in probability.
