Probability

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Probability — Quick Facts

Key Definitions:

• Experiment: Any action or process that leads to one or more outcomes
• Sample space (S): The set of all possible outcomes
• Event (E): A subset of the sample space
• Probability of an event: $P(E) = \frac{n(E)}{n(S)}$ where $0 \leq P(E) \leq 1$

Basic Probability Rules:

• $P(S) = 1$ (certain event)
• $P(\emptyset) = 0$ (impossible event)
• $P(E’) = 1 - P(E)$ (complement rule)
• $P(E \cup F) = P(E) + P(F) - P(E \cap F)$ (addition rule)

For Equally Likely Outcomes: $$P(\text{event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$$

Venn Diagram Formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

If A and B are mutually exclusive (cannot happen together): $$P(A \cup B) = P(A) + P(B)$$

⚡ JAMB Exam Tip: If the question uses words like “or,” “either,” or “at least one,” use the addition rule. If it uses “and,” “both,” or “simultaneously,” you may need the multiplication rule.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Probability — Study Guide

Conditional Probability:

$P(A|B)$ = probability of A given that B has occurred

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

Multiplication Rule: $$P(A \cap B) = P(A) \times P(B|A) = P(B) \times P(A|B)$$

Independent Events: Events A and B are independent if $P(A \cap B) = P(A) \times P(B)$

Example: Tossing a coin twice — the result of the first toss doesn’t affect the second.

Worked Example:

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.

$P(\text{first red}) = \frac{5}{8}$ $P(\text{second red | first red}) = \frac{4}{7}$ $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

Worked Example with Complement:

In a class of 40 students, 25 study Physics, 30 study Chemistry, and 5 study neither. A student is chosen at random. What is the probability that they study both subjects?

$n(P) = 25$, $n(C) = 30$, $n(\text{neither}) = 5$ $n(\text{both}) = 25 + 30 + 5 - 40 = 20$ (using inclusion-exclusion) $P(\text{both}) = \frac{20}{40} = \frac{1}{2}$

⚡ Common Student Mistake: For “without replacement” problems, remember to reduce the sample size after each draw. For “with replacement,” the probability stays the same for each draw.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Probability — Comprehensive Notes

Bayes’ Theorem:

$$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$$

Extended Bayes’ Theorem (for multiple mutually exclusive events): $$P(A_i|B) = \frac{P(B|A_i) \cdot P(A_i)}{\sum_{j=1}^{n} P(B|A_j) \cdot P(A_j)}$$

Worked Example:

An exam has three questions. Question 1: 80% of students answer correctly. Question 2: 70% answer correctly. Question 3: 60% answer correctly. Assuming questions are independent, find the probability that a student gets all three wrong.

$P(\text{all wrong}) = P(\text{wrong Q1}) \times P(\text{wrong Q2}) \times P(\text{wrong Q3})$ $= 0.20 \times 0.30 \times 0.40 = 0.024 = 2.4%$

Permutations and Combinations in Probability:

When selecting $r$ items from $n$ items:

• Without replacement, order matters (permutation): $P(n, r) = \frac{n!}{(n-r)!}$
• Without replacement, order doesn’t matter (combination): $C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Worked Example:

From a standard deck of 52 cards, 3 cards are dealt. Find the probability of getting exactly 2 hearts.

Number of ways to choose 2 hearts: $C(13, 2) = 78$ Number of ways to choose 1 non-heart: $C(39, 1) = 39$ Total favourable: $78 \times 39 = 3042$

Total ways to choose 3 cards: $C(52, 3) = 22100$

$P(\text{2 hearts}) = \frac{3042}{22100} = \frac{1521}{11050} \approx 0.138$

Expectation (Expected Value):

If a random variable $X$ takes values $x_1, x_2, …, x_n$ with probabilities $p_1, p_2, …, p_n$: $$E(X) = \sum_{i=1}^{n} x_i \cdot p_i$$

Example: A game costs ₦50 to play. You win ₦150 with probability 0.2, ₦50 with probability 0.5, and lose otherwise. Find expected profit.

$E(\text{profit}) = (150-50)(0.2) + (50-50)(0.5) + (-50)(0.3)$ $= 100(0.2) + 0(0.5) - 50(0.3)$ $= 20 - 15 = ₦5$

Since $E(\text{profit}) > 0$, the game is favourable on average.

Probability Distribution:

$x$	$P(X=x)$
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

Check: $\sum P(X=x) = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$ ✓

JAMB Pattern Analysis (2015-2024):

• 2015: Simple probability with cards and dice
• 2017: Conditional probability with Venn diagrams
• 2019: Binomial probability (exactly $k$ successes in $n$ trials)
• 2021: Expected value word problem
• 2023: Multiplication rule with dependent events

⚡ Exam Strategy: Draw a tree diagram for multi-step probability problems. Label every branch with its probability. Multiply along branches for “AND” events, add for “OR” events at a given node.