Further Mathematics · Lesson 18

Probability: laws, conditional & tree diagrams

WAECNECO

What the syllabus asks

WAEC: Statistics and probability: sample space, laws of probability (addition and multiplication), mutually exclusive, independent and exhaustive events, conditional probability, tree diagrams, Bayes-type reverse probability

NECO: Statistics and probability: sample space, laws of probability (addition and multiplication), mutually exclusive, independent and exhaustive events, conditional probability, tree diagrams, Bayes-type reverse probability

What you will learn

List sample spaces and compute simple probabilities.
Use the complement rule, especially for "at least one".
Apply the addition law, distinguishing mutually exclusive events.
Apply the multiplication law for independent and dependent events.
Compute conditional probabilities from tables, Venn diagrams and formulas.
Draw and use tree diagrams, including reverse (Bayes-type) questions.

Key points

P(E) = favourable ÷ total, for equally likely outcomes; 0 ≤ P ≤ 1.
P(not A) = 1 − P(A); "at least one" = 1 − P(none).
P(A ∪ B) = P(A) + P(B) − P(A ∩ B); mutually exclusive → no overlap.
Independent: P(A ∩ B) = P(A)P(B). Dependent: P(A ∩ B) = P(A)P(B | A).
Without replacement: change denominators for later draws.
Tree diagrams: multiply along, add between; tips sum to 1.
P(A | B) = P(A ∩ B)/P(B); restrict to the given row, column or set.
Reverse probability: P(cause | result) = P(cause and result)/P(result).

Start here: Sample spaces and basic probability

Probability measures how likely something is, on a scale from 0 (impossible) to 1 (certain). When all outcomes are equally likely:

P(E) = (number of outcomes in E) ÷ (total number of outcomes in the sample space)

The sample space is the full list of outcomes. Listing it carefully is the foundation of every probability problem. For one die it is {1, 2, 3, 4, 5, 6}. For two coins it is {HH, HT, TH, TT}: note HT and TH are different outcomes. For two dice there are 6 × 6 = 36 outcomes, best shown in a grid.

Worked example 18.1

Two fair dice are thrown. Find the probability that the total is 7.

Sample space: 36 equally likely pairs.
Pairs giving 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1): 6 pairs.
P = 6/36.

1/6

Worked example 18.2

A bag holds 4 green, 6 yellow and 2 white balls. One is picked at random. Find P(yellow) and P(not white).

Total 12. P(yellow) = 6/12 = 1/2.
P(white) = 2/12 = 1/6, so P(not white) = 1 − 1/6 = 5/6.

1/2; 5/6

Experimental probability (relative frequency) = number of times the event happened ÷ number of trials. If a tossed drawing pin lands point-up 36 times in 120 tosses, the estimated probability is 36/120 = 0.3. More trials give a better estimate.

⚠️ Common mistake: treating "a total of 7" and "a total of 2" as equally likely. Only (1,1) gives 2, so P(2) = 1/36.

Try it yourself. Two coins are tossed. Find P(exactly one head).

Show solution

HT, TH out of 4: 1/2.

Listing outcomes systematically

For two stages, draw a grid with the first stage down the side and the second along the top. For a coin and a die there are 2 × 6 = 12 outcomes: H1, H2, …, H6, T1, …, T6. For three coins there are 2³ = 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. A tidy list protects you from missing outcomes. For example, P(exactly two heads in three tosses) = 3/8, because three of the eight outcomes (HHT, HTH, THH) have exactly two heads. If you had listed only "two heads and one tail" as a single outcome, you would wrongly get 1/4. The rule is that the outcomes you count must be equally likely.

A useful check: probabilities of all outcomes in a sample space add up to exactly 1. For the two-dice grid, the totals 2, 3, …, 12 have frequencies 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, which add up to 36.

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Practice questions

1. Two dice are thrown. Find P(total is 10 or more).

Show worked answer

(4,6),(5,5),(6,4),(5,6),(6,5),(6,6): 6/36 = 1/6.

2. P(A) = 0.4, P(B) = 0.5, independent. Find P(A ∪ B).

Show worked answer

0.4 + 0.5 − 0.2 = 0.7.

3. A box has 6 good and 4 bad oranges. Two are taken without replacement. Find P(one good, one bad).

Show worked answer

2 × 6/10 × 4/9 = 48/90 = 8/15.

4. P(a seed germinates) = 0.8. Find P(at least one of 3 germinates).

Show worked answer

1 − 0.2³ = 0.992.

5. P(A ∩ B) = 0.15, P(A) = 0.6. Find P(B | A).

Show worked answer

0.15/0.6 = 0.25.

Key words

Experiment: An action with uncertain outcomes, e.g. tossing a coin.
Sample space: The set of all possible outcomes.
Event: A set of outcomes from the sample space.
Mutually exclusive: Events that cannot happen together.
Independent: Events where one does not affect the probability of the other.
Exhaustive: Events that together cover the whole sample space.
Complement: The event "not A", written A′.
Conditional probability: The probability of A given that B has happened, P(A|B).
Tree diagram: A branching diagram showing successive outcomes and their probabilities.
With replacement: The item is returned before the next draw, so probabilities do not change.
Without replacement: The item is not returned, so later probabilities change.
Reverse (Bayes) probability: Finding the probability of a cause given an observed result.

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