How to 'AND' and 'OR' probabilites? Consider this question: A marksman fires at a target consisting of a central disk, and two concentrating rings. The probability of hitting the disk and ring are, $0.35, 0.3, 0.25$ respectively. What is the probability of missing the target?
Let,
$D$ = Event that marksman hit the disk
$R_1$ = Event that marksman hit the Ring 1
$R_2$= Event that marksman hit the Ring 2
Therefore, $$P(D)=0.35, P(R_1)=0.3, P(R_2)=0.25$$
I tried to solve it in two different ways:
AND way: $$P(\text{miss})=P(\text{miss disk AND miss Ring 1 AND miss Ring 2})\\=P(\text{miss disk})P(\text{miss ring 1})P(\text{miss ring 2})\\=(1-0.35)(1-0.3)(1-0.25)=0.34$$
OR way: $$P(\text{miss})=1-P(\text{hit})\\=1-P(\text{hit disk OR hit Ring 1 OR hit Ring 2})\\=1-(0.35+0.3+0.25)=0.1$$
What mistake am I doing? Which one is the correct answer?
Ever since I started doing Computer Science, {AND, multiplication, intersection, conjunction, min} feels like they are all same in my mind. And so is, {OR, addition, union, disjunction, max}.
So, for mutually excusive events, AND-ing probability is just multiplying them and OR-ing is just adding them.  And since the hitting of disks and rings in my marksman example are mutually exclusive (can't hit disk and ring at same time), I should get same answer either way.
Please help me understand AND-ing and OR-ing probabilities of two events.

$P(A \mbox{ AND } B) = P(A) \cdot P(B)$ if $A$ and $B$ are independent.
$P(A \mbox{ OR } B) = P(A) + P(B)$ if they are mutually exclusive.
These are two very different conditions; in fact, independent events can never be mutually exclusive (or vica versa) unless one of the events is impossible.
In your case, "hit disk", "hit ring 1", "hit ring 2" are mutually exclusive, because they correspond to disjoint points on the target.  So the "OR way" is correct.
You cannot solve this with the "AND way" because (say) "miss disk" and "miss ring 1" are not independent: Given that you've missed the disk, then you have a higher probability of hitting ring 1 than you started with.
