Mutually Exclusive Events A and B

What Does Mutually Exclusive Mean?

Imagine you have two choices, but you can only pick one, not both at the same time. Well, in the world of probability, this is similar to the concept of Mutually Exclusive Events (other cool names: Disjoint Events).

Two events (let's call them event A and event B) are said to be mutually exclusive if both events cannot possibly occur at the same time in a single trial. Simply put, if A happens, B cannot happen. If B happens, A cannot happen.

Characteristics of Mutually Exclusive Events

The main characteristic is just that: cannot happen simultaneously. There's no outcome that can belong to event A and event B at the same time.

Probability of Event A AND B Occurring Together

Because events A and B cannot happen together if they are mutually exclusive, the probability of both occurring simultaneously is zero!

We can write the probability of event "A and B" (both occurring) as:

Or using the intersection symbol:

Remember, if they are mutually exclusive, their intersection is empty, so the probability is zero!

Calculating the Probability of A OR B for Mutually Exclusive Events

So, how do we calculate the probability of event A happening OR event B happening if A and B are mutually exclusive?

Since they can't happen together, we simply add the probabilities of each individual event.

The formula becomes very easy:

Or using the union symbol:

This is the Special Addition Rule which applies ONLY to mutually exclusive events. (If the events are not mutually exclusive, there's a slightly different formula).

Examples of Mutually Exclusive Events

To understand better, look at these examples:

1. Coin Toss:

   The event of getting "Heads" and the event of getting "Tails". They can't happen together, right?

   • $P(\text{Heads}) = 1/2$
   • $P(\text{Tails}) = 1/2$
   • Probability of getting Heads OR Tails = $P(\text{Heads}) + P(\text{Tails}) = 1/2 + 1/2 = 1$ (One of them must occur).

2. Rolling a Die (once):

   • The event of getting a "3" and the event of getting a "5". You can't get both numbers at once.
     • $P(3) = 1/6$
     • $P(5) = 1/6$
     • Probability of getting a 3 OR 5 = $P(3) + P(5) = 1/6 + 1/6 = 2/6 = 1/3$.
   • The event of getting an "even number" ($\{2, 4, 6\}$) and the event of getting an "odd number" ($\{1, 3, 5\}$). No number is both even and odd.
     • $P(\text{Even}) = 3/6 = 1/2$
     • $P(\text{Odd}) = 3/6 = 1/2$
     • Probability of getting Even OR Odd = $P(\text{Even}) + P(\text{Odd}) = 1/2 + 1/2 = 1$.

3. Drawing a Card (once):

   • The event of getting a "King" and the event of getting a "Queen". One card cannot be both a King and a Queen.
     • There are 4 Kings in 52 cards, $P(\text{King}) = 4/52$
     • There are 4 Queens in 52 cards, $P(\text{Queen}) = 4/52$
     • Probability of getting a King OR Queen = $P(\text{King}) + P(\text{Queen}) = 4/52 + 4/52 = 8/52 = 2/13$.
   • The event of getting a "Red card (Hearts/Diamonds)" and the event of getting a "Club ($\clubsuit$)". Clubs are black, so a card cannot be red and a club simultaneously.
     • There are 26 red cards, $P(\text{Red}) = 26/52$
     • There are 13 clubs, $P(\clubsuit) = 13/52$
     • Probability of getting Red OR Club = $P(\text{Red}) + P(\clubsuit) = 26/52 + 13/52 = 39/52 = 3/4$.