Exhaustive and Mutually Exclusive Events in Probability

Exhaustive and Mutually Exclusive Events

To solve probability problems effectively for the ESAT, you must distinguish between two key properties of events: exhaustion and mutual exclusivity.

A set of events is defined as exhaustive if at least one of the events in the set must occur. In any experiment, the exhaustive set covers all possible outcomes. For example, if you roll a standard die, the events 'rolling a number less than 5' and 'rolling a number greater than 3' are exhaustive because every possible result ($1, 2, 3, 4, 5, 6$) is contained in at least one of those categories.

Events are mutually exclusive if no more than one of the events can occur at the same time. For instance, if you roll a die, the events 'landing on a 2' and 'landing on a 5' are mutually exclusive because a single roll cannot produce both results simultaneously.

It is common for a set of events to be both exhaustive and mutually exclusive. Consider rolling a fair six-sided die again: the events 'land on an even number' and 'land on an odd number' are both exhaustive (as at least one must occur) and mutually exclusive (as a number cannot be both even and odd).

The probabilities of a set of events that are both exhaustive and mutually exclusive sum to $1$. This fact is a powerful tool for finding unknown probabilities.

The Probability of an Event Not Occurring

Two events, $A$ and 'not $A$', always form a set that is both exhaustive and mutually exclusive. This leads to the fundamental identity:

$P(A) + P(\text{not } A) = 1$

This equation allows us to find the probability of an event not occurring if we know the probability of the event occurring, or vice-versa.

Worked Examples: Complementary Events

Example 1: Playing Cards If a card is picked at random from a standard deck, the probability of it being an ace is $\frac{1}{13}$. What is the probability that it is not an ace?

$P(\text{ace}) + P(\text{not ace}) = 1$ $P(\text{not ace}) = 1 - \frac{1}{13} = \frac{12}{13}$

Example 2: Weather Forecasts The probability of rain tomorrow is $0.3$. What is the probability of no rain tomorrow?

$P(\text{rain}) + P(\text{not rain}) = 1$ $P(\text{not rain}) = 1 - 0.3 = 0.7$

Example 3: Book Types Marta picks a book at random from a library. The probability that the book is a fiction book is $65\%$. What is the probability that the book is non-fiction?

$P(\text{fiction}) + P(\text{not fiction}) = 100\%$ $P(\text{not fiction}) = 100\% - 65\% = 35\%$

Finding an Unknown Probability

When multiple outcomes are mutually exclusive and exhaustive, we can determine a single missing probability by subtracting the sum of the known probabilities from $1$.

Worked Example: Coloured Counters A bag contains only red counters, green counters, and yellow counters. A counter is picked at random. The probability of picking a red counter is $0.35$ and the probability of picking a yellow counter is $0.2$. Work out the probability of picking a green counter.

Since the bag contains only these three colours, the events are exhaustive. Since a counter cannot be more than one colour, they are mutually exclusive. Therefore:

$P(\text{red}) + P(\text{green}) + P(\text{yellow}) = 1$ $0.35 + P(\text{green}) + 0.2 = 1$ $P(\text{green}) = 1 - 0.35 - 0.2 = 0.45$

Finding an Unknown Probability with Ratios

More advanced problems require combining the sum-to-one property with ratios.

Worked Example: The Biased Spinner Marisa has a biased spinner that can land on red, green, blue, yellow, or purple. The probability that it lands on red is $\frac{1}{5}$, on blue is $\frac{1}{8}$, and on yellow is $\frac{3}{10}$. The ratio of the probability of landing on green to the probability of landing on purple is $3 : 2$. Work out the probability of the spinner landing on green.

1. State the sum: $P(\text{red}) + P(\text{green}) + P(\text{blue}) + P(\text{yellow}) + P(\text{purple}) = 1$

2. Substitute known values and find a common denominator: We use $40$ as the common denominator for the known fractions. $\frac{8}{40} + P(\text{green}) + \frac{5}{40} + \frac{12}{40} + P(\text{purple}) = 1$ $\frac{25}{40} + P(\text{green}) + P(\text{purple}) = 1$

3. Find the combined remaining probability: $P(\text{green}) + P(\text{purple}) = 1 - \frac{25}{40} = \frac{15}{40}$

4. Divide by the ratio: The ratio $3 : 2$ has $3 + 2 = 5$ parts. One part is $\frac{15}{40} \div 5 = \frac{3}{40}$.

5. Calculate the specific outcome: Since green corresponds to $3$ parts of the ratio: $P(\text{green}) = 3 \times \frac{3}{40} = \frac{9}{40}$