Logic behind "OR"? $P(\bar{A}\cap B) + P(\bar{B}\cap A)$ Vs $P(A \cup B)$ In a deck of $52$ cards , find the probability of drawing a king or a black card .
Let probability of drawing a black card be P(A)
Let probability of drawing a king be P(B)
As drawing king , drawing black card are not mutually exclusive ,
by "OR" I mean english word or in the following explanation.
I thought taking a king of spade or club will violate the condition black "OR" king thinking that when I take a king of spade or club will fulfil both the conditions. so I did as 
P(A OR B) = P(A) + P(B) - 2 P(A$\cap$B) 
= $ \frac{4}{52} + \frac{26}{52} - 2 * \frac{2}{52}$
= $ \frac{26}{52}$
But the book told as per conventions as per the formula 
P(A OR B) = P(A) + P(B) - P(A$\cap$B) = $28/52$
Why is my reasoning wrong ?
when I draw a black spade or black club , won't I violate the condition that it is black "OR" king ? so only I subtracted possibilities of drawing black spade or black club . so I subtracted $2/52$ and got $26/52$ instead of $28/52$.
If I can also apply my reasoning as
P(black or king) = P(King that is not black) + P(Black card that is not a king)
= $2/52 + 24/52 = 26/52 $
Please help me in knowing where I am going wrong in reasoning 
So I am saying the formula to be used here is $P(\bar{A}\cap B) + P(\bar{B}\cap A)$ and book says the formula to be used here is $P(A  \cup B)$. So in my logic for OR it is real A or B , there is not option for A and B.
My formula for for this question -> $P(\bar{A}\cap B) + P(\bar{B}\cap A)$  = $P(A) + P(B) -2\times P(A \cap B)$
Book's formula(conventional) for this question ->  $P(A  \cup B)$ = $P(A) + P(B) - 1 \times P(A \cap B)$
 A: Usually the OR operator is understood in the inclusive sense, that is A OR B is true if either A is true, B is true, or both are true. Therefore
$$ P(A \cup B) = P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) = P(A) + P(B) - P(A \cap B), $$
rather than the formula you wrote.
A: I think that formula you are using is wrong. It should be:
P(A OR B) = P(A) + P(B) - P(A$\cap$B), which will give the true answer. I don't get it why we would exclude the number twice? We only need to exclude just once, beacuse the king of spades and the king of clubs were included twice.

First try to make things clearer to the reader. OR means the same as $\lor$, which means the either one of the condition is met or both ar met. But it seems that you are interested in exclusive OR (XOR) which is usually represented by $\oplus$. The formula for this is:
$$P(A \oplus B) = P(A \cup B) \cap \overline{P(A \cap B)}$$
But your formula would do the trick. 
This formula first add the probability of either of the condition is fulfiled and then excludes the probability when both od them are met. And your formula add the probability of just one condition separately.
But anyway you'll get the same result.
A: This an English problem, not a mathematics problem. 
The OR operator in Boolean logic is defined by: A  OR  B  is true if A is true , or B is true , or both are true.  The truth table for a OR-gate reflects this.
The XOR, exclusive-OR operator in Boolean logic is defined by: A  XOR  B  is true if A is true , or B is true , but not if  both are true.  The truth table for a XOR-gate reflects this.
So the question is:  did the writer of the question have the OR type of or,  or the XOR type of or in mind?
Consider this:  If you were told that you could be charged if you were speeding or driving drunk, would you even consider the defense that you were doing both, and therefore immune to charges?
