Conditional Probability as a Restricted Sample Space

Why Restrict the Sample Space?

A probability question can change when some information is already known. Suppose a fair die is rolled, and someone tells you that the result is even. The possible results are no longer all six faces: given that information, only 2, 4, and 6 remain relevant. If you now ask for the probability that the result is greater than 3, only 4 and 6 of those three possible outcomes meet the condition.

This is the central idea of conditional probability: once an event is known to have occurred, calculate probabilities using that event as the restricted sample space. The key is to count the outcomes that meet both the condition and the event of interest, then compare that count with the number of outcomes in the restricted space.

Definition: Probability Within a Restricted Space

Let $A$ and $B$ be events. The event $B$ supplies the information that is known to hold, and $A$ is the event whose probability we want after taking that information into account. Outcomes that belong to both $A$ and $B$ are written $A\cap B$.

Definition: If $P(B)>0$, the conditional probability of $A$ given $B$ is $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. $$ In a finite sample space where all outcomes are equally likely, this is the number of outcomes in $B$ that also belong to $A$, divided by the total number of outcomes in $B$.

The denominator is $P(B)$, not the probability of the whole sample space. Knowing that $B$ occurred means we restrict attention to $B$. The numerator counts the outcomes in that restricted space that also satisfy $A$. If all outcomes are equally likely, we can make the same calculation by counting rather than by first calculating probabilities.

The condition $P(B)>0$ matters. If $B$ is impossible, there is no possible outcome in the restricted space, so the ratio in the definition has a zero denominator. In that case $P(A\mid B)$ is not defined by this formula.

Why the Counting Formula Works

Consider a finite sample space with $N$ equally likely outcomes. Suppose $B$ contains $m$ outcomes, where $m>0$. Once we are told that $B$ occurred, those $m$ outcomes are the possibilities we compare. Of these, the outcomes that also satisfy $A$ are exactly the outcomes in $A\cap B$. Therefore the conditional probability is the fraction of the $m$ restricted outcomes that belong to $A\cap B$.

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

Here $|C|$ denotes the number of outcomes in a finite event $C$. To see that this agrees with the probability-ratio definition, let $k=|A\cap B|$. Because the original $N$ outcomes are equally likely, $P(A\cap B)=k/N$ and $P(B)=m/N$. Thus

\frac{P(A\cap B)}{P(B)} = \frac{k/N}{m/N} = \frac{k}{m} = \frac{|A\cap B|}{|B|}.

The original sample space provides the starting point, but once $B$ is given, the relevant denominator is the size or probability of $B$. The formula follows from using the restricted set of possible outcomes as the new reference space.

Worked Example: An Even Result on a Fair Die

Hypothetical experiment. A fair six-sided die is rolled once. Given that the result is even, find the probability that it is greater than 3.

The original sample space is $\{1,2,3,4,5,6\}$. Since the die is fair, these six outcomes are equally likely. Let $B$ be the event that the result is even, and let $A$ be the event that the result is greater than 3.

The known information restricts the sample space to $B=\{2,4,6\}$, which has three outcomes. The outcomes in this restricted set that also satisfy $A$ are $A\cap B=\{4,6\}$, which has two outcomes. Therefore

P(A\mid B)=\frac{|A\cap B|}{|B|} =\frac{2}{3} \approx 0.667.

So, given that the die shows an even number, the probability that it shows a number greater than 3 is about 0.667. The denominator is 3 because only the three even results remain relevant; using all six die faces would ignore the information in the question.

Worked Example: Two Coin Tosses

Hypothetical experiment. Two fair coins are tossed, and the outcomes are recorded in order. Given that the first toss is heads, find the probability of getting exactly one head.

For two ordered tosses, the equally likely outcomes are $HH$, $HT$, $TH$, and $TT$. Let $B$ be the event that the first toss is heads, so $B=\{HH,HT\}$. Let $A$ be the event that exactly one of the two tosses is heads, so $A=\{HT,TH\}$.

The restricted space $B$ contains two outcomes. Only $HT$ belongs to both $A$ and $B$, so $A\cap B=\{HT\}$ contains one outcome. The conditional probability is

P(A\mid B)=\frac{|A\cap B|}{|B|} =\frac{1}{2} =0.500.

Thus, among the outcomes in which the first toss is heads, half have exactly one head. The outcome $TH$ does have exactly one head, but it is not in the restricted space because its first toss is tails. It cannot be counted after the condition is imposed.

Worked Example: Selecting a Number from 1 to 12

Hypothetical experiment. One integer is chosen at random from 1 through 12, with every integer equally likely. Given that the selected number is a multiple of 3, find the probability that it is even.

Let $B$ be the event that the number is a multiple of 3. Then $B=\{3,6,9,12\}$, so the restricted space has four outcomes. Let $A$ be the event that the number is even. The outcomes satisfying both conditions are $A\cap B=\{6,12\}$, giving two favorable outcomes.

Using the count within the restricted space,

P(A\mid B)=\frac{|A\cap B|}{|B|} =\frac{2}{4} =0.500.

The probability is 0.500: among the four multiples of 3, two are even. Notice that the total of 12 possible integers is not the denominator after we know the selected number is a multiple of 3.

Using the Probability-Ratio Definition

Counting is convenient when outcomes are equally likely. The definition using event probabilities also applies when outcomes are not equally likely, provided $P(B)>0$. In such a setting, take the probability of the outcomes that satisfy both events and divide it by the probability of the condition.

For example, suppose a hypothetical spinner has four regions with different areas, so the regions are not equally likely. Let $B$ be the event that the spinner lands in a particular set of regions, and let $A$ be another event. Counting regions would not generally give the right probability: a small region and a large region do not contribute equal chances. The probability-ratio definition still uses $P(A\cap B)/P(B)$, with probabilities that account for the spinner’s unequal regions.

In the examples above, each elementary outcome was equally likely, so the event-probability ratio reduced to a simple count ratio. Always check whether equal likelihood has been established before using raw counts as probabilities.

Common Mistakes

Keeping the original total as the denominator. A learner might say that two die results are even and greater than 3, so the answer is $2/6$. This uses all six faces, despite being told the result is even. The correction is to restrict the sample space to $\{2,4,6\}$, giving $2/3$.
Counting outcomes that meet only one condition. In the two-coin example, $TH$ has exactly one head, but it does not belong to the restricted event that the first toss is heads. Count only outcomes that satisfy both $A$ and $B$.
Assuming a condition makes outcomes equally likely. Restricting a sample space does not make unequal-probability outcomes equal. For non-equally likely outcomes, use $P(A\cap B)/P(B)$, not a ratio of the number of regions or labels.
Ignoring an impossible condition. If $P(B)=0$, the defining ratio has denominator zero. Do not calculate a conditional probability by dividing by zero.

Key takeaway: To find the probability of $A$ given $B$, treat $B$ as the relevant sample space. In an equally likely finite space, divide the number of outcomes in both $A$ and $B$ by the number in $B$. In general, divide $P(A\cap B)$ by $P(B)$, provided $P(B)>0$.

Check Your Understanding

For each question, identify the restricted event before calculating.

A fair die is rolled. Given that the result is odd, what is the probability that it is greater than 3? List the restricted outcomes and the outcomes meeting both conditions.
Two fair coins are tossed in order. Given that the second toss is tails, what is the probability of exactly one head?
An integer is selected uniformly from 1 through 10. Given that it is greater than 5, what is the probability that it is even?
In $P(A\mid B)=P(A\cap B)/P(B)$, explain what the numerator and denominator represent.
Why is the count ratio not generally valid if the outcomes in the original sample space are not equally likely?