Mutually Exclusive and Exhaustive Events
This topic covers the fundamental rule that the total probability of all possible outcomes in a situation must add up to 1. This principle is a key tool for finding an unknown probability when all other probabilities are known.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The sum of the probabilities for every possible outcome of an experiment is always exactly 1 (or 100%).
- This rule applies to sets of events that are both 'exhaustive' (cover all possibilities) and 'mutually exclusive' (cannot happen at the same time).
- A common application is finding the probability of an event NOT happening: P(not A) = 1 - P(A).
- If you know the probabilities of all but one event in an exhaustive, mutually exclusive set, you can find the last one by subtracting the sum of the others from 1.
Formulae
P(A) + P(B) + ⋯ + P(N) = 1 When A, B, ⋯, N form a complete set of mutually exclusive and exhaustive events.
P(not A) = 1 - P(A) To find the probability of an event not occurring, given the probability that it does occur.
Definitions
- Exhaustive Events
- A set of events that covers all possible outcomes. At least one of these events is guaranteed to occur.
- Mutually Exclusive Events
- Events that cannot happen simultaneously. If one event occurs, it excludes the possibility of the others occurring.
Worked example
A bag contains only red, blue, and green marbles. The probability of picking a blue marble is 1/3. The probability of picking a green marble is twice the probability of picking a red marble. What is the probability of picking a green marble?
- 1
The events 'picking red', 'picking blue', and 'picking green' are exhaustive and mutually exclusive, so their probabilities sum to 1:
P(R) + P(B) + P(G) = 1.
- 2
Substitute the known probability for blue:
P(R) + 1/3 + P(G) = 1.
- 3
Isolate the sum of the unknown probabilities:
P(R) + P(G) = 1 - 1/3 = 2/3 - 4
Use the relationship given:
P(G) = 2 × P(R) - 5
Substitute this into the equation from the previous step:
P(R) + 2*P(R) = 2/3.
- 6
Simplify and solve for P(R):
3 × P(R) = 2/3, which means P(R) = (2/3) / 3 = 2/9 - 7
Finally, calculate P(G) using the relationship:
P(G) = 2 × P(R) = 2 × (2/9) = 4/9
Answer: 4/9
Common mistakes
- ×Making arithmetic errors when adding or subtracting multiple fractions, especially when finding a common denominator.
- ×Incorrectly subtracting decimals from 1, for instance calculating 1 - 0.45 as 0.65 instead of 0.55.
- ×Failing to correctly set up an equation when probabilities are linked by a ratio or a statement like 'twice as likely'.
No-calculator tips
- ✓When subtracting multiple fractions from 1, add the fractions together first before performing a single subtraction. For example, to find 1 - 1/5 - 1/8, calculate 1/5 + 1/8 = 8/40 + 5/40 = 13/40, then do 1 - 13/40 = 27/40.
- ✓Convert percentages and decimals to fractions to simplify calculations, especially if other probabilities are already fractions. For example, 0.25 becomes 1/4.
- ✓To subtract a decimal from 1, think in terms of 'complements to 100'. For example, for 1 - 0.35, think 'what do I add to 35 to get 100?'. The answer is 65, so the result is 0.65.