Conditional Probability on a Restricted Sample Space

What Changes When an Event Is Known?

Suppose a report says that a randomly selected claim came from a certain group. If we then ask for the probability that the claim exceeded a threshold, we are no longer considering every claim in the original model. We are considering only claims from that group. Conditional probability formalizes this change: it restricts attention to an event known to have occurred and measures probabilities relative to that restricted sample space.

The key is to distinguish the event that is known to have occurred from the event whose probability is being asked for. The restricted event determines which outcomes remain possible; the target event determines which of those remaining outcomes count as favorable.

Definition: For events $A$ and $B$ in a probability model, if $P(B)>0$, the conditional probability of $A$ given $B$ is $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. $$ The event $B$ is the conditioning event, and $A\cap B$ consists of outcomes where both $A$ and $B$ occur.

The denominator is the probability of the restricted sample space. The numerator is the probability of the part of that restricted space that also satisfies the target event. Because the definition requires division by $P(B)$, it applies only when $P(B)>0$. In this tutorial, conditioning always means conditioning on an event with positive probability.

Restrict First, Then Renormalize

In a finite model, imagine keeping only the outcomes in $B$. The original probabilities of those outcomes may add to less than one, because outcomes outside $B$ have been removed. To make the probabilities within the restricted model add to one, divide each remaining outcome's probability by $P(B)$. Then add the resulting probabilities for outcomes that also belong to $A$.

This gives a direct justification of the definition. In a finite sample space, write $p(\omega)$ for the probability of outcome $\omega$. The probability assigned to an outcome $\omega$ after restricting to $B$ is $p(\omega)/P(B)$, for $\omega\in B$. Thus the conditional probability of $A$ is the sum of those adjusted probabilities over outcomes in $A\cap B$:

P(A\mid B) =\sum_{\omega\in A\cap B}\frac{p(\omega)}{P(B)} =\frac{\sum_{\omega\in A\cap B}p(\omega)}{P(B)} =\frac{P(A\cap B)}{P(B)}.

If all outcomes in the original sample space are equally likely, the common probability assigned to each outcome cancels from the numerator and denominator. Conditional probability can then be calculated as a count of favorable outcomes within $B$, divided by the total count within $B$. This count shortcut is not valid when the outcomes are not equally likely.

Worked Example: Doubles Given a Large Dice Total

A hypothetical game rolls two fair six-sided dice. The outcomes are ordered pairs, so $(2,5)$ and $(5,2)$ are distinct. Given that the sum is at least 9, find the probability that the roll is a double. Use the restricted-sample-space method and identify the counting trap.

Model and events. There are $6\times 6=36$ equally likely ordered pairs. Let $B$ be the event that the sum is at least 9, and let $A$ be the event that the two dice show the same number.

Count the restricted sample space. The ordered pairs with sums 9, 10, 11, and 12 are, respectively, 4, 3, 2, and 1 outcomes. Therefore, $B$ has $4+3+2+1=10$ outcomes, so $P(B)=10/36$.

Count outcomes satisfying both events. A double has the form $(k,k)$. For the sum to be at least 9, $2k\geq 9$, so $k$ must be 5 or 6. The qualifying doubles are $(5,5)$ and $(6,6)$, giving $P(A\cap B)=2/36$.

Apply the definition.

P(A\mid B)=\frac{2/36}{10/36}=\frac{2}{10}=\frac{1}{5}=0.2000.

Check. Among the 10 ordered outcomes in the restricted space, exactly 2 are doubles. The result is $2/10=1/5$, as calculated. The common mistake is to treat the four possible sums, 9 through 12, as equally likely and calculate $2/4=1/2$. They are not equally likely: those sums contain 4, 3, 2, and 1 ordered outcomes. The method avoids that mistake by counting the original equally likely outcomes, not the sum categories.

Unequal Probabilities: Keep the Weights

When outcomes have different probabilities, restricting the sample space does not make the remaining outcomes equally likely. Instead, retain their original relative weights and divide by the total probability of the conditioning event. This is the same definition, but the calculation must use probabilities rather than a simple count.

Worked Example: A Weighted Selector and an Even Result

A hypothetical selector chooses one of the numbers 1 through 6, with probabilities $1/21, 2/21, 3/21, 4/21, 5/21,$ and $6/21$, respectively. Given that the result is at least 3, find the probability that it is even. State why counting the remaining values alone would be wrong.

Model and events. The listed probabilities sum to $(1+2+3+4+5+6)/21=21/21=1$, so they define the model. Let $B$ mean that the result is at least 3, and let $A$ mean that the result is even.

Find the probability of the condition. The outcomes in $B$ are 3, 4, 5, and 6. Their probability weights are 3, 4, 5, and 6, so

P(B)=\frac{3+4+5+6}{21}=\frac{18}{21}.

Find the probability of both events. The outcomes that are at least 3 and even are 4 and 6. Therefore,

P(A\cap B)=\frac{4+6}{21}=\frac{10}{21}.

Restrict and renormalize.

P(A\mid B)=\frac{10/21}{18/21} =\frac{10}{18} =\frac{5}{9} \approx 0.5556.

Check. Within the restricted outcomes, the total weight is 18 and the even outcomes carry weight 10. The conditional probability is therefore $10/18$. Counting two even results among four remaining values would give $2/4$, which is incorrect because the four values do not have equal probabilities.

Continuous Models: Restrict the Region

For a continuous model, outcomes are points or values in a region rather than a short list. If the model is uniform over a region, probability is proportional to length, area, or volume, as appropriate. Conditional probability can then be calculated by taking the size of the part satisfying both events relative to the size of the region satisfying the conditioning event. For a nonuniform continuous model, probability is determined by the density over each region; area alone is not enough.

The ratio still uses probabilities, not just geometric sizes. In a uniform model, the common density factor cancels, leaving a ratio of areas. The conditioning region must have positive probability so the denominator is nonzero.

Worked Example: A Point in a Square Given a Triangular Condition

A point $(X,Y)$ is selected uniformly from the unit square $0\leq X\leq 1,\ 0\leq Y\leq 1$. Given that $X+Y<1$, find the probability that $X<1/2$. Use areas to restrict the model, and verify the total area used in the calculation.

Model and events. The square has area 1 and uniform density 1, so the probability of any region in the square equals its area. Let $B$ be the triangular region $X+Y<1$, and let $A$ be the region $X<1/2$.

Find the area of the conditioning region. For each $x$ from 0 to 1, the allowed $y$-values in $B$ run from 0 to $1-x$. Thus,

P(B)=\int_0^1(1-x)\,dx =\left[x-\frac{x^2}{2}\right]_0^1 =1-\frac{1}{2} =\frac{1}{2}.

Find the area satisfying both events. In $A\cap B$, $x$ runs from 0 to $1/2$, and for each such $x$, $y$ runs from 0 to $1-x$. Hence,

P(A\cap B)=\int_0^{1/2}(1-x)\,dx =\left[x-\frac{x^2}{2}\right]_0^{1/2} =\frac{1}{2}-\frac{1}{8} =\frac{3}{8}.

Calculate the conditional probability.

P(A\mid B)=\frac{3/8}{1/2}=\frac{3}{4}=0.7500.

Check. The conditioning triangle has area $1/2$, and the part of it with $X<1/2$ has area $3/8$, so the conditional share is $(3/8)/(1/2)=3/4$. The common mistake is to divide the intersection area by the area of the whole square, obtaining $3/8$. That is $P(A\cap B)$, not the probability of $A$ within the restricted region $B$.

What the Definition Does Not Cover

The condition $P(B)>0$ matters. For example, in a continuous model, the probability that a continuously distributed variable equals one exact value is often zero. The ratio in the definition would then have a zero denominator, so this event-based definition cannot be used to condition on that exact value. Conditioning on a positive-probability interval or region is different: its probability can be calculated and used as the denominator, as in the square example.

Also keep the roles of the two events straight. In $P(A\mid B)$, $B$ is what is known to have occurred, so it supplies the denominator. The numerator must include only outcomes that are both in $B$ and in the target event $A$. Reversing the order changes the question and generally changes the answer.

Key takeaway: To calculate $P(A\mid B)$, first verify that $P(B)>0$. Then restrict attention to $B$, find the probability of the part also in $A$, and divide by $P(B)$. Use counts only when the relevant outcomes are equally likely.

Check Your Understanding

For each problem, identify the restricted event and calculate the target probability within that restriction.

A fair six-sided die is rolled. Given that the result is greater than 2, find the probability that it is divisible by 3. (Answer: $1/2$)
One card is drawn uniformly from a standard 52-card deck. Given that the card is red, find the probability that it is a face card (jack, queen, or king). (Answer: $3/13$)
A selector returns 1 through 5 with probabilities proportional to the returned number. Given that its result is at least 2, find the probability that it is odd. (Answer: $4/7$)
A value $X$ is selected uniformly from the interval $[0,8]$. Given that $X>2$, find the probability that $X\leq 5$. (Answer: $1/2$)
A point is selected uniformly from the unit square. Given that $X+Y<1$, find the probability that $X>1/2$. (Answer: $1/4$)

What You'll Learn