Probability Basics: Events, Outcomes and Rules

Probability is the mathematics of uncertainty. It lets us put a number on how likely something is, from rolling a six to a medicine working. Once you understand a handful of key terms and rules, you can solve the vast majority of probability questions in Matric, FSc, O/A Level and first-year university courses.

Key Terms

• Experiment: a process with uncertain results, such as tossing a coin or rolling a die.
• Outcome: a single possible result, such as getting a 4.
• Sample space \( S \): the set of all possible outcomes. For one die, \( S = \{1, 2, 3, 4, 5, 6\} \).
• Event: a set of outcomes we are interested in, such as "an even number" \( = \{2, 4, 6\} \).

The Classical Definition

When all outcomes are equally likely, the probability of an event \( A \) is

$$ P(A) = \frac{n(A)}{n(S)} = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}} $$

For example, the probability of rolling an even number is \( \frac{3}{6} = \frac{1}{2} \).

Basic Facts

• \( 0 \le P(A) \le 1 \) for every event.
• \( P(S) = 1 \): something in the sample space must happen.
• \( P(\varnothing) = 0 \): an impossible event has probability zero.

The Main Rules

1. Complement Rule

The complement \( A' \) means "\( A \) does not happen":

$$ P(A') = 1 - P(A) $$

This is especially useful for "at least one" questions.

2. Addition Rule

For any two events,

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$

We subtract \( P(A \cap B) \) because outcomes in both events would otherwise be counted twice.

If \( A \) and \( B \) are mutually exclusive (they cannot happen together), then \( P(A \cap B) = 0 \) and the rule simplifies to \( P(A \cup B) = P(A) + P(B) \).

3. Multiplication Rule for Independent Events

Two events are independent if one occurring does not affect the probability of the other. Then

$$ P(A \cap B) = P(A) \times P(B) $$

Examples include separate coin tosses or separate dice rolls. (For dependent events you need conditional probability, covered in a separate guide.)

Mutually Exclusive vs Independent

These are often confused, but they are very different:

• Mutually exclusive: cannot happen together, so \( P(A \cap B) = 0 \).
• Independent: do not influence each other, so \( P(A \cap B) = P(A)P(B) \).

Two events with non-zero probabilities cannot be both mutually exclusive and independent.

Worked Example 1: Two Dice

Question: Two fair dice are rolled. Find the probability that (a) the sum is 7, (b) the sum is 7 or both dice show the same number.

Step 1: Sample space. Each die has 6 outcomes, so \( n(S) = 6 \times 6 = 36 \) equally likely ordered pairs.

Step 2: Part (a). The pairs that sum to 7 are

$$ (1,6),\; (2,5),\; (3,4),\; (4,3),\; (5,2),\; (6,1) $$

That is 6 outcomes, so

$$ P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6} $$

Step 3: Part (b). "Doubles" are \( (1,1), (2,2), \dots, (6,6) \): 6 outcomes, so \( P(\text{double}) = \frac{6}{36} \).

Can a double sum to 7? That would need two equal numbers adding to 7, which is impossible since 7 is odd. So the events are mutually exclusive:

$$ P(\text{sum } 7 \text{ or double}) = \frac{6}{36} + \frac{6}{36} = \frac{12}{36} = \frac{1}{3} $$

Worked Example 2: Cards and the Addition Rule

Question: A card is drawn from a standard 52-card deck. Find the probability that it is a king or a heart.

Step 1: Individual probabilities.

• Kings: 4 cards, so \( P(K) = \frac{4}{52} \).
• Hearts: 13 cards, so \( P(H) = \frac{13}{52} \).

Step 2: Overlap. One card is both a king and a heart (the king of hearts), so \( P(K \cap H) = \frac{1}{52} \).

Step 3: Apply the addition rule.

$$ P(K \cup H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} $$

Check by counting: 13 hearts plus the 3 other kings gives 16 cards. Correct.

Worked Example 3: "At Least One" with the Complement

Question: A fair die is rolled three times. What is the probability of getting at least one six?

Step 1: Use the complement. "At least one six" is the complement of "no sixes".

Step 2: Probability of no six on one roll. \( \frac{5}{6} \).

Step 3: Independent rolls, so multiply.

$$ P(\text{no sixes}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} $$

Step 4: Subtract from 1.

$$ P(\text{at least one six}) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421 $$

Listing every case directly would be far longer; the complement makes it a three-line calculation.

Common Mistakes

• Adding probabilities of overlapping events without subtracting the intersection.
• Multiplying probabilities of dependent events as if they were independent, for example drawing cards without replacement.
• Treating outcomes as equally likely when they are not. The sums 2 to 12 on two dice are not equally likely; the ordered pairs are.
• Forgetting order matters when listing dice outcomes: \( (2,5) \) and \( (5,2) \) are different.
• Getting a probability above 1 and not noticing that something has gone wrong.

Tips

• Draw a table, tree diagram or Venn diagram before calculating.
• Look for the words "or" (addition rule), "and" (multiplication rule) and "at least" (complement rule).
• Simplify fractions at the end, and check that your answer lies between 0 and 1.

Key Takeaways

• Probability of an event = favourable outcomes ÷ total outcomes, when outcomes are equally likely.
• Complement rule: \( P(A') = 1 - P(A) \), ideal for "at least one" questions.
• Addition rule: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
• Multiplication rule for independent events: \( P(A \cap B) = P(A)P(B) \).
• Mutually exclusive and independent are different ideas; do not confuse them.

If probability questions keep tripping you up, send them to Amathify on WhatsApp or through the "Ask a Question" form, and our experts and PhD scholars will solve them step by step for you.