Probability and Permutations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Probability measures how likely an event is to occur. It is expressed as a number between 0 and 1, where 0 means impossible and 1 means certain. In NECO mathematics, you must master both basic probability concepts and counting principles (permutations and combinations).

Key Definitions

• Experiment: An action with uncertain outcomes, such as tossing a coin or rolling a die.
• Outcome: A single possible result of an experiment. “Landing on heads” is one outcome of a coin toss.
• Sample space: The set of all possible outcomes. For a six-sided die, $S = {1, 2, 3, 4, 5, 6}$.
• Event: A specific outcome or set of outcomes we are interested in. “Rolling an even number” = ${2, 4, 6}$.

Probability Formula

$$P(\text{event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$$

Example: What is the probability of rolling a 3 on a fair six-sided die? $$P(3) = \frac{1}{6}$$

The Complementary Event

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

If $P(\text{passing an exam}) = 0.7$, then $P(\text{failing}) = 1 - 0.7 = 0.3$.

Types of Probability

Type	Meaning
Theoretical	Based on logical analysis (fair coin: $P(H) = 1/2$)
Experimental	Based on actual observations (survey results)
Conditional	Probability given that another event has occurred

⚡ NECO Exam Tips

• Always check whether events are mutually exclusive before adding probabilities. $P(A \text{ or } B) = P(A) + P(B)$ only if $A$ and $B$ cannot happen together.
• In “and” probability questions for independent events, multiply: $P(A \text{ and } B) = P(A) \times P(B)$.
• Read carefully: “at least one” often means use complementary probability: $1 - P(\text{none})$.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Permutations and Combinations

These are two different ways of selecting items from a set.

Fundamental Counting Principle

If task 1 can be done in $m$ ways and task 2 can be done in $n$ ways, both tasks can be done in $m \times n$ ways.

Example: A student has 3 shirts and 2 pairs of trousers. How many different outfits can be worn? $3 \times 2 = 6$ outfits.

Permutations (Order Matters)

$$^nP_r = \frac{n!}{(n-r)!}$$

Where $n!$ (n factorial) = $n \times (n-1) \times (n-2) \times … \times 2 \times 1$

Example: How many ways can 3 students be chosen from 5 to sit in 3 specific chairs? $$^5P_3 = \frac{5!}{(5-3)!} = \frac{120}{2!} = \frac{120}{2} = 60$$

Another example: Arrange the letters of the word “EXAM”:

• Total arrangements = $4! = 24$
• Vowels in vowels positions: 2 E, A → arrangements = $2! = 2$
• Consonants in consonant positions: X, M → arrangements = $2! = 2$
• Total = $2 \times 2 = 4$

Combinations (Order Does Not Matter)

$$^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Example: A committee of 3 is to be formed from 7 students. How many ways? $$\binom{7}{3} = \frac{7!}{3!4!} = \frac{5040}{6 \times 24} = \frac{5040}{144} = 35$$

Permutation vs Combination Quick Reference

Question asks…	Use
Arrange, seat, line up, order	Permutation
Select, choose, form a committee, group	Combination

Probability of Combined Events

Mutually Exclusive Events ($A \cap B = \emptyset$): $$P(A \text{ or } B) = P(A) + P(B)$$

Non-Mutually Exclusive Events: $$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

Example: One card is drawn from a standard deck. Find $P(\text{King or Heart})$. $P(\text{King}) = 4/52$ $P(\text{Heart}) = 13/52$ $P(\text{King of Hearts}) = 1/52$ $P(\text{King or Heart}) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13$

Independent Events ($P(A \text{ and } B) = P(A) \times P(B)$): $$P(A \text{ and } B) = P(A) \times P(B)$$

Example: Toss a fair coin twice. Find $P(\text{head both times})$. $P(H \text{ and } H) = 1/2 \times 1/2 = 1/4$

Conditional Probability

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

Example: A bag contains 3 red and 5 blue balls. Two balls are drawn without replacement. Find $P(\text{second ball is red | first ball is blue})$.

$P(\text{first blue}) = 5/8$ After removing one blue: $P(\text{second red}) = 3/7$ $P(\text{second red | first blue}) = 3/7$

Common Mistakes

1. Confusing permutations and combinations — always ask: does order matter?
2. Forgetting to divide by factorials in combination formulas
3. In probability questions with “without replacement”, remember the sample space changes after each draw.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Advanced Probability and Counting Techniques

Bayes’ Theorem

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

Example Problem: A test for a disease is 99% accurate. If 1% of the population has the disease, what is the probability that a person who tests positive actually has the disease?

Let $D$ = has disease, $T^+$ = tests positive

$P(D) = 0.01$ $P(T^+|D) = 0.99$ $P(T^+|\text{not } D) = 0.01$ (false positive) $P(\text{not } D) = 0.99$

$P(T^+) = P(T^+|D) \times P(D) + P(T^+|\text{not } D) \times P(\text{not } D)$ $P(T^+) = 0.99 \times 0.01 + 0.01 \times 0.99 = 0.0099 + 0.0099 = 0.0198$

$P(D|T^+) = \frac{0.99 \times 0.01}{0.0198} = \frac{0.0099}{0.0198} = 0.5$

So even with a 99% accurate test, if you test positive and only 1% of the population has the disease, there’s a 50% chance you actually have it!

Permutations with Repeated Elements

The number of arrangements of $n$ objects where some are identical: $$\frac{n!}{n_1! \times n_2! \times …}$$

Example: Arrange the letters of “STATISTICS”: Total letters = 10 S appears 3 times, T appears 3 times, I appears 2 times $$\frac{10!}{3! \times 3! \times 2!} = \frac{3628800}{6 \times 6 \times 2} = \frac{3628800}{72} = 50400$$

Circular Permutations

Number of ways to arrange $n$ objects in a circle = $(n-1)!$

Example: In how many ways can 6 people be seated at a round table? $(6-1)! = 5! = 120$ ways

Note: For arrangements around a circular table where clockwise and anticlockwise are considered the same, divide by 2 again.

The Binomial Distribution

For repeated independent trials with probability $p$ of success: $$P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$$

Example: A fair coin is tossed 5 times. Find $P(\text{exactly 3 heads})$.

$n = 5$, $r = 3$, $p = 0.5$ $$P(X=3) = \binom{5}{3} \times (0.5)^3 \times (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125$$

Expected Value (Mathematical Expectation)

$$E(X) = \sum x \cdot P(x)$$

Example: A game costs #100 to play. You win #300 if you roll a 6 on a fair die, and #50 otherwise. Is the game fair?

$E(X) = 300 \times (1/6) + 50 \times (5/6) - 100 = 50 + 41.67 - 100 = -8.33$

The game is not fair — on average, you lose #8.33 per game.

Combinatorial Proofs

Sometimes combinations can be used to prove identities: $$\binom{n}{r} = \binom{n}{n-r}$$

This makes intuitive sense: choosing $r$ from $n$ is the same as leaving out $n-r$.

NECO SSCE Past Question Patterns

Probability questions in NECO typically:

• Test basic probability formula and complementary events (very common)
• Include permutations and combinations in Section B
• Ask about sample spaces and events (drawing cards, throwing dice)
• Include “at least one” problems solved via complementary probability

Typical marks distribution: Probability 8–12 marks, Permutations 6–10 marks.

Key Formulas Summary

$$P(A) = \frac{n(A)}{n(S)}$$ $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ $$P(A \cap B) = P(A) \times P(B) \text{ (independent)}$$ $$^nP_r = \frac{n!}{(n-r)!}$$ $$^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

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