Theorem of the Sum of Mutually Exclusive Events
If E1 and E2 are mutually exclusive events, the probability of their union—meaning the probability that at least one of E1 or E2 occurs—is equal to the sum of the probabilities of the individual events: $$ P(E1∪E2)=P(E1)+P(E2) $$
Where
- P(E1∪E2) is the probability of the union, meaning the probability that at least one of E1 or E2 occurs.
- P(E1) is the probability that event E1 occurs.
P(E2) is the probability that event E2 occurs.
The theorem of the sum of mutually exclusive events is a fundamental principle of probability theory.
When are two events mutually exclusive?
Two events are called mutually exclusive (or disjoint) when the occurrence of one automatically excludes the possibility of the other happening in the same experiment.
More formally, two events E1 and E2 are mutually exclusive if their intersection is an empty set.
$$ P(E1∩E2)=Ø $$
Proof. The proof of this theorem follows directly from the classical definition of probability, where the probability of an event is the ratio of favorable outcomes to the total number of possible outcomes. In the case of mutually exclusive events, since there is no overlap, the favorable outcomes for the union correspond to the sum of the favorable outcomes of the individual events.
A Practical Example
An urn contains 20 balls: 8 black, 7 red, and 5 white.
If I draw a ball at random from the urn, there are three possible outcomes:
- E1 = drawing a black ball
- E2 = drawing a red ball
- E3 = drawing a white ball
These are mutually exclusive events because the occurrence of one excludes the others.
The probabilities of these events are:
$$ p(E_1) = \frac{8}{20} = \frac{4}{10} = \frac{2}{5} $$
$$ p(E_2) = \frac{7}{20} $$
$$ p(E_3) = \frac{5}{20} = \frac{1}{4} $$
Therefore, the probability of drawing either a black or white ball is:
$$ p(E_1 \cup E_3) = \frac{2}{5} + \frac{1}{4} $$
$$ p(E_1 \cup E_3) = \frac{2 \cdot 4 + 1 \cdot 5}{20} $$
$$ p(E_1 \cup E_3) = \frac{8 + 5}{20} $$
$$ p(E_1 \cup E_3) = \frac{13}{20} $$
$$ p(E_1 \cup E_3) = 0.65 $$
So, the probability of drawing either a black or white ball is 0.65, or 65%.
Example 2
Consider a deck of 52 playing cards.
E1: "Drawing an Ace"
E2: "Drawing the King of Hearts"
These two events are mutually exclusive because you can't draw both an Ace and the King of Hearts with a single card.
Since there are four Aces in a deck, the probability of drawing an Ace (E1) is 4/52:
$$ p(E1) = \frac{4}{52} = \frac{1}{13} $$
The probability of E2 is 1/52 because there is only one King of Hearts in the deck.
$$ p(E2) = \frac{1}{52} $$
Using the theorem of the sum of mutually exclusive events, we can find the probability of drawing either an Ace or the King of Hearts:
$$ P(E1∪E2)= P(E1) + P(E2) $$
$$ P(E1∪E2)= \frac{1}{52}+ \frac{4}{52} $$
$$ P(E1∪E2)= \frac{1+4}{52} $$
$$ P(E1∪E2)= \frac{5}{52} $$
So, the probability of drawing either an Ace or the King of Hearts from a standard deck is 5/52, which is approximately 9.62%.
And so on.
