# 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.

$P(A \mbox{ AND } B) = P(A) \cdot P(B)$ 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.

For the AND way, you can only multiply probabilities if you know they are independent. Since the events appear to be mutually exclusive, we know they aren't independent: Hitting the disk makes hitting a ring impossible, and similarly missing the disk makes hitting a ring more likely and thus missing a ring less likely (you know you didn't land in the ring region).

The correct approach is the OR one.

The experiment has one trial (one shot) and four outcomes (hit disk, hit ring 1, hit ring 2, miss). Since the four trials are mutually exclusive and collectively exhaustive, then: $$P(D)+P(R_1)+P(R_2)+P(M)=1 \Rightarrow P(M)=1-0.35-0.3-0.25=0.1.$$ If you want to use "AND" operation, then use addition rule: $$P(M)=P(D^C\cap R_1^C\cap R_2^C)=P(D^C\cup R_1^C\cup R_2^C)-P(D^C)-P(R_1^C)-P(R_2^C)+P(D^C\cap R_1^C)+P(D^C\cap R_2^C)+P(R_1^C\cap R_2^C)=1-0.65-0.7-0.75+0.35+0.4+0.45=0.1.$$

What you found in "AND" case was an experiment with three trials (three shots) and four outcomes. Namely $P(D^C)P(R_1^C)P(R_2^C)$ shows the probability of missing disk, then ring 1, then ring 2 in three independent shots.