Exhaustive and Mutually Exclusive Events in Probability

Exhaustive Sets of Outcomes

In probability theory, an exhaustive set of outcomes refers to a collection of all possible results of an experiment. Because these outcomes represent every possibility, it is certain that exactly one of them will occur in any given trial.

For any experiment, the total probability of all possible outcomes must be $1$. This is the mathematical representation of certainty. If we have a set of $n$ distinct outcomes $O_1, O_2, \dots, O_n$, the property states that:

$P(O_1) + P(O_2) + \dots + P(O_n) = 1$

Mutually Exclusive Events

Two events are mutually exclusive if they cannot happen at the same time. For example, when rolling a single die, the event 'rolling a 2' and the event 'rolling a 5' are mutually exclusive because a single die cannot show both numbers simultaneously. Mathematically, for mutually exclusive events $A$ and $B$, the probability of both occurring is $P(A \cap B) = 0$.

When a set of events is both mutually exclusive and exhaustive, they partition the entire sample space into non-overlapping sections that account for every possibility.

The Sum to One Property

If we are given a complete list of all mutually exclusive events for an experiment, we can use the property that their probabilities sum to one to solve for unknown variables. This is a common technique in TMUA questions involving biased coins, spinners, or selection from bags.

Worked Example: Finding an Unknown Probability

A biased four sided spinner is marked with the numbers 1, 2, 3, and 4. The probabilities of the spinner landing on 1, 2, and 3 are given as $P(1) = 0.15$, $P(2) = 0.25$, and $P(3) = 0.4$. Find the probability that the spinner lands on 4.

Solution:

1. Identify that the outcomes {1, 2, 3, 4} are exhaustive (they are the only outcomes mentioned) and mutually exclusive (the spinner only lands on one number).
2. Set up the equation: $P(1) + P(2) + P(3) + P(4) = 1$.
3. Substitute the known values: $0.15 + 0.25 + 0.4 + P(4) = 1$.
4. Simplify: $0.8 + P(4) = 1$.
5. Solve for $P(4)$: $P(4) = 1 - 0.8 = 0.2$.

Algebraic Applications

TMUA questions often involve algebra to test the depth of your understanding of this property. You may be required to solve for a constant $k$ that defines a probability distribution.

Worked Example: Solving for a Constant

A bag contains red, blue, and yellow balls. The probability of picking a red ball is $2k$, the probability of picking a blue ball is $k$, and the probability of picking a yellow ball is $k^2$. Given these are the only colours in the bag, find the possible values of $k$.

Solution:

1. Since these are the only colours, the events are exhaustive. They are also mutually exclusive as only one ball is picked.
2. Sum the probabilities to one: $k^2 + 2k + k = 1$.
3. Form a quadratic equation: $k^2 + 3k - 1 = 0$.
4. Use the quadratic formula to solve for $k$: $k = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)} = \frac{-3 \pm \sqrt{13}}{2}$.
5. Since probability must be non-negative, $k$ cannot be $\frac{-3 - \sqrt{13}}{2}$ (which is negative). Therefore, the only valid solution is $k = \frac{-3 + \sqrt{13}}{2}$.

The Complement Rule

A direct consequence of the exhaustive property is the Complement Rule. For any event $A$, the events $A$ and 'not $A$' (denoted $A'$) are mutually exclusive and exhaustive. Therefore:

$P(A) + P(A') = 1 \implies P(A') = 1 - P(A)$

This is often the most efficient way to solve problems asking for the probability of 'at least one' outcome, by subtracting the probability of 'none' from 1.