Conditional Probability (Higher)

[Higher tier] Everything in this lesson is Higher-tier content. So far, repeated events have been independent — replacing a counter kept the probabilities the same. But what happens when an item is not replaced? The bag changes, and the probability of the second draw depends on the first. This is conditional probability: the probability of an event given that another event has already happened. This lesson covers dependent events, drawing tree diagrams without replacement, the conditional probability notation $P(A \mid B)$ , and the conditional formula.

This is firmly AO1 (calculating conditional probabilities and building dependent trees), AO2 (recognising dependence) and AO3 (multi-stage "without replacement" reasoning). OCR command words include Work out, Calculate, Show that and Complete.

Key Vocabulary

Term	Definition
Dependent events	Events where the outcome of one changes the probability of the other.
Without replacement	An item is not put back, so the total (and possibly the count) changes.
Conditional probability	The probability of $A$ given that $B$ has happened, written $P(A \mid B)$ .
$P(A \mid B)$	Read as "the probability of $A$ given $B$ ".
Reduced sample space	The smaller set of outcomes once the condition $B$ is known.

Dependent Events and "Without Replacement"

When an item is drawn and not replaced, the second draw has a smaller total, and if the item removed was of the type you are tracking, that count drops too. The two draws are then dependent.

This is the single most important idea in the lesson, so it is worth slowing down. Imagine a bag of $8$ counters, $5$ of them red. On the first draw, $P(\text{red}) = \dfrac{5}{8}$ . If you draw a red and do not put it back, the bag now holds only $7$ counters, of which $4$ are red — so the probability of red on the second draw has changed to $\dfrac{4}{7}$ . Both numbers moved: the denominator fell from $8$ to $7$ (one fewer counter overall), and the numerator fell from $5$ to $4$ (one fewer red). If instead you had drawn a blue first, the second-draw probability of red would be $\dfrac{5}{7}$ — the same $5$ reds, but now out of $7$ . So the second probability depends on what happened first, which is exactly what "dependent" means.

Contrast this with the with replacement case from the previous lesson, where you put the counter back, the bag returns to $8$ counters with $5$ red, and the probability stays $\dfrac{5}{8}$ every time. The single phrase "without replacement" in a question is your signal to reduce the totals on the second stage; "with replacement" (or "the counter is returned") is your signal to keep them the same.

Worked Example 1

A bag contains $5$ red and $3$ blue counters. Two counters are drawn without replacement. Work out the probability that both are red.

Solution: First draw: $P(\text{red}) = \dfrac{5}{8}$ . After removing a red, $4$ reds remain out of $7$ counters, so the second probability is $\dfrac{4}{7}$ .

$P(\text{R and R}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$

Common error: Keeping the second probability as $\dfrac{5}{8}$ . Without replacement, both the count of reds and the total change.

Worked Example 2

From the same bag ( $5$ red, $3$ blue), two counters are drawn without replacement. Work out $P(\text{both blue})$ .

Solution: $P(\text{B and B}) = \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}$

Tree Diagrams Without Replacement

The tree-diagram method is unchanged, but the second-stage probabilities differ depending on the first outcome. This is the only thing that changes from the with-replacement case: you still multiply along paths and add between paths, but the second column of branches is no longer a copy of the first. Each second-stage branch must be worked out for its own situation — what the bag looks like after the particular first draw on that path. Here is the tree for the bag of $5$ red and $3$ blue counters, drawn without replacement.

graph LR
    S["Start"] -->|"5/8"| R1["Red (1st)"]
    S -->|"3/8"| B1["Blue (1st)"]
    R1 -->|"4/7"| RR["Red, Red"]
    R1 -->|"3/7"| RB["Red, Blue"]
    B1 -->|"5/7"| BR["Blue, Red"]
    B1 -->|"2/7"| BB["Blue, Blue"]

Notice each pair of second-stage branches still sums to $1$ (for example $\dfrac{4}{7} + \dfrac{3}{7} = 1$ ), but the values differ between the "after red" and "after blue" branches — that is conditional probability in action.

Worked Example 3

Using the tree above, work out the probability of getting one counter of each colour.

Solution: "One of each" is Red,Blue or Blue,Red.

$P(\text{R,B}) = \dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}, \qquad P(\text{B,R}) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$

$P(\text{one of each}) = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28}$

Worked Example 4

A box has $4$ toffees and $6$ mints. Two sweets are taken without replacement. Work out the probability that at least one is a toffee.

Solution: Use the complement. $P(\text{no toffee}) = P(\text{mint then mint}) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$ .

$P(\text{at least one toffee}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

Conditional Probability Notation $P(A \mid B)$

The symbol $P(A \mid B)$ means "the probability of $A$ given that $B$ has happened". Knowing $B$ restricts attention to a reduced sample space.

The "reduced sample space" idea is the key to every conditional question. Once you are told that $B$ has happened, the outcomes where $B$ did not happen are no longer possible, so they drop out of consideration. You are now working within a smaller world — only the outcomes consistent with $B$ — and you ask what fraction of those also satisfy $A$ . This is why the denominator of a conditional probability is the number in $B$ , not the original grand total. A frequent and costly error is to keep dividing by the original total; the whole point of "given $B$ " is that the total has shrunk to the size of $B$ .

The bar symbol " $\mid$ " is read aloud as "given". So $P(\text{rain} \mid \text{cloudy})$ is "the probability of rain given that it is cloudy", and it is generally different from $P(\text{rain})$ on its own, because knowing it is cloudy changes how likely rain becomes. Conditional probability is precisely the mathematics of how information updates a probability. One subtlety to watch: $P(A \mid B)$ and $P(B \mid A)$ are usually different — "the probability of rain given clouds" is not the same as "the probability of clouds given rain" — so always read carefully which event is the condition (the one after the bar) and which is the event of interest.

Worked Example 5

A bag has $6$ red and $4$ green counters. Two are drawn without replacement. Work out $P(\text{second is green} \mid \text{first is red})$ .

Solution: Given the first counter was red, the bag now holds $5$ red and $4$ green — $9$ counters.

$P(\text{second green} \mid \text{first red}) = \dfrac{4}{9}$

This is exactly a second-stage branch on a tree diagram: the branch value is a conditional probability.

Worked Example 6

Of $30$ students, $18$ study French and, of those, $7$ also study Spanish. A student who studies French is chosen at random. Work out the probability they also study Spanish.

Solution: The condition restricts us to the $18$ French students; $7$ of them study Spanish.

$P(\text{Spanish} \mid \text{French}) = \dfrac{7}{18}$

The Conditional Probability Formula

The formal definition is

$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$

In words: divide the probability that both happen by the probability of the condition. This formula is just the reduced-sample-space idea written with probabilities instead of counts. Restricting to the outcomes where $B$ happens means dividing by $P(B)$ ; asking how many of those also satisfy $A$ means the top is $P(A \text{ and } B)$ . If you prefer counting, the same statement reads $P(A \mid B) = \dfrac{n(A \text{ and } B)}{n(B)}$ , which is exactly what you do when reading a conditional probability off a two-way table or a Venn diagram.

There is a second, equally useful, form of the formula. Multiplying both sides by $P(B)$ gives

$P(A \text{ and } B) = P(B) \times P(A \mid B)$

Conditional Probability (Higher)

Conditional Probability (Higher)

Key Vocabulary

Dependent Events and "Without Replacement"

Worked Example 1

Worked Example 2

Tree Diagrams Without Replacement

Worked Example 3

Worked Example 4

Conditional Probability Notation P(A∣B)P(A \mid B)P(A∣B)

Worked Example 5

Worked Example 6

The Conditional Probability Formula

More in Mathematics

Conditional Probability Notation $P(A \mid B)$