IMAT Mathematics · Probability and Statistics

Probability: Outcomes, Rules and Trees

Probability is favourable outcomes over equally likely outcomes; the addition rule handles 'or', multiplication handles 'and', and the second draw changes when nothing is put back.

All Probability and Statistics notes

Why this matters

Each ministry paper with a probability item used two dice (2023) or two balls drawn from a bag, with replacement (2024) or without (2026). An older question moved an object from one box to another before a second draw, which is a two-stage tree.

Concept 1 of 4: Probability from equally likely outcomes, and the complement

When every outcome is equally likely, probability is a fraction: how many outcomes you want out of how many there are. With two dice it pays to picture the 6 by 6 grid of 36 ordered pairs, because a 2 then a 5 is a different outcome from a 5 then a 2. When the event you want is messy, count what you do not want and subtract from 1.

Definition

For equally likely outcomes: $P(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of outcomes}}$ .
$0 \le P(A) \le 1$ : 0 means impossible, 1 means certain.
Complement: $P(\text{not } A) = 1 - P(A)$ . Use it for "at least one".
Two dice give 36 equally likely ordered pairs; two coins give 4 (HH, HT, TH, TT).

Equally likely outcomes

P(A) = \frac{n(A)}{n(S)} \qquad P(\text{not } A) = 1 - P(A)

$n(A)$ number of outcomes in the event
$n(S)$ number of possible outcomes

Worked example

Two fair dice are rolled. Find the probability that the total is 9, and the probability that at least one die shows a 6.

Practice this conceptself-check · 4 quick reps

With two dice, (2, 5) and (5, 2) are different outcomes

The 36 ordered pairs are equally likely; the 21 unordered pairs are not (a double can happen one way, a mixed pair two ways). Count ordered pairs, so a mixed pair counts twice and a double once.

Concept 2 of 4: The addition rule for 'A or B'

Adding

P(A)

and

P(B)

counts the overlap, where both happen, twice, so you take it away once. Picture two overlapping circles in a box: the region covered by either circle is the two circles minus the lens they share. If the circles do not overlap, nothing is subtracted.

Definition

Addition rule: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ .
Mutually exclusive events cannot happen together: $P(A \text{ and } B) = 0$ , so the probabilities simply add.
"Neither A nor B" is the complement of "A or B": $1 - P(A \text{ or } B)$ .

Addition rule

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

$A \cup B$ A or B (or both)
$A \cap B$ A and B together

Worked example

A card is drawn from a standard pack of 52. What is the probability that it is a heart or a king?

Practice this conceptself-check · 4 quick reps

Subtract the overlap when events can happen together

P(A) + P(B)

alone is right only for mutually exclusive events. If the sum comes out above 1, or the events clearly overlap (a heart and a king), subtract

P(A \text{ and } B)

Concept 3 of 4: Independent events and tree diagrams (with replacement)

Two events are independent when one happening does not change the chance of the other: a coin has no memory. Then the chance of both is the product. A tree diagram draws each stage as branches; you multiply along a path and add the paths that give the result you want.

Definition

A and B are independent if $P(A \text{ and } B) = P(A) \times P(B)$ .
Repeated draws with replacement (the object is put back) are independent: every draw has the same probabilities.
Tree diagram: one set of branches per stage, each labelled with its probability; the branches from one point add up to 1.
Multiply along a path for "this and then that"; add separate paths for "this way or that way".
"Exactly one success in $n$ tries" has $n$ paths, one for each position of the success.

Independent events

P(A \cap B) = P(A)\,P(B)

Worked example

A basketball player scores each free throw with probability 0.8, independently. She takes two. Find the probability that she scores both, exactly one, and at least one.

Practice this conceptself-check · 4 quick reps

Exactly one needs every order

"Exactly one red in three spins" is three different paths on the tree. Multiplying the probabilities along a single path gives only one of them; multiply by the number of orders.

Concept 4 of 4: Conditional probability and drawing without replacement

Once you know something has happened, you look only at the cases where it happened. Drawing without replacement is the everyday example: after one ball leaves the bag there is one ball fewer, and one fewer of its colour, so the second draw has new probabilities.

Definition

$P(B \mid A)$ , read "the probability of B given A", is the probability of B when A is known to have happened.
$P(B \mid A) = \dfrac{P(A \text{ and } B)}{P(A)}$ , so $P(A \text{ and } B) = P(A) \times P(B \mid A)$ .
Without replacement: the second draw's denominator is one less, and the numerator is one less if the first draw took that colour.
From a two-way table, $P(B \mid A)$ is the count in both A and B divided by the count in A.
If A and B are independent, $P(B \mid A) = P(B)$ . In general $P(B \mid A) \ne P(A \mid B)$ .

Conditional probability

P(B \mid A) = \frac{P(A \cap B)}{P(A)}

Worked example

A bag holds 4 blue and 6 yellow marbles. Two are taken without replacement. Find the probability that both are blue, and that one of each colour is taken.

Practice this conceptself-check · 4 quick reps

Without replacement, the second fraction changes

After one draw the bag has one item fewer, so the second denominator drops by one. Keeping the same fraction for both draws answers the with-replacement question instead, and that answer is always among the options.

P(B given A) is not P(A given B)

The probability that a biology student studies chemistry is not the probability that a chemistry student studies biology: the two divide the same overlap by different totals.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Probability from equally likely outcomes, and the complement
Equally likely outcomes
$P(A) = \frac{n(A)}{n(S)} \qquad P(\text{not } A) = 1 - P(A)$
The addition rule for 'A or B'
Addition rule
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Independent events and tree diagrams (with replacement)
Independent events
$P(A \cap B) = P(A)\,P(B)$
Conditional probability and drawing without replacement
Conditional probability
$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$

Watch out for (5)

With two dice, (2, 5) and (5, 2) are different outcomes
→ Probability from equally likely outcomes, and the complement
Subtract the overlap when events can happen together
→ The addition rule for 'A or B'
Exactly one needs every order
→ Independent events and tree diagrams (with replacement)
Without replacement, the second fraction changes
→ Conditional probability and drawing without replacement
P(B given A) is not P(A given B)
→ Conditional probability and drawing without replacement

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