Counting Outcomes Under Conditional Restrictions

Why the Denominator Changes

A finite probability problem may describe a random selection or arrangement and then reveal that a particular restriction holds. The requested probability is now conditional: the outcomes that violate the given restriction are no longer possible for the calculation. The key exam skill is to count the restricted sample space correctly and use its size—not the size of the original sample space—as the denominator.

For example, if a committee is selected uniformly from all committees of a fixed size, each committee is an equally likely outcome. If we are told that the committee contains at least one person from a specified group, we consider only committees satisfying that condition. We then count how many of those also satisfy the event being asked about.

Definition: Let $\Omega$ be a finite set of equally likely outcomes, let $A$ be the event described by the given condition, and let $B$ be the event of interest. If $P(A)>0$, then $$ P(B\mid A)=\frac{|A\cap B|}{|A|}. $$ Here, $|A|$ is the number of outcomes satisfying the condition, and $|A\cap B|$ is the number satisfying both the condition and the event of interest.

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

The unrestricted total $|\Omega|$ cancels. This justification depends on equal likelihood of the original outcomes. If outcomes are not equally likely, counting them is not enough: conditional probability must be calculated from their probabilities, using $P(B\mid A)=P(A\cap B)/P(A)$.

Choose the Right Kind of Count

Before counting, decide whether an outcome records an unordered selection or an ordered result. If the order of $r$ selected objects from $n$ distinct objects does not matter, the number of selections is $\binom{n}{r}$. If the order matters and no object can be repeated, the number of ordered arrangements is $n!/(n-r)!$. A shuffled row of distinct objects generally has $n!$ possible arrangements.

Then translate both the condition and the target event into countable restrictions. It often helps to count the denominator and numerator separately. A complement can simplify a count: for instance, “at least one red” may be counted as “all possible outcomes minus outcomes with no red.”

Specify the elementary outcomes.
Decide exactly what counts as one outcome, including whether order matters and whether repetition is allowed.

Count the outcomes satisfying the condition.
This restricted total is the denominator. Apply complements or cases if they make the count clearer.

Count the outcomes satisfying both requirements.
Keep the original condition in force while adding the target event.

Divide and check.
Compute the conditional probability and verify that the numerator counts a subset of the denominator.

Worked Examples

Worked Example: A Committee With a Group Restriction

A training unit has 3 calibration specialists and 4 field specialists. A committee of 4 distinct people is selected uniformly at random. Given that the committee includes at least one calibration specialist, find the probability that it includes exactly 2 calibration specialists.

Method. Count unordered committees. The condition is “at least one calibration specialist,” so count all committees and subtract those with none.

There are

\binom{7}{4}=35

committees in total. A committee with no calibration specialist must consist of all 4 field specialists, so there is

\binom{4}{4}=1

such committee. The restricted denominator is $35-1=34$.

For the target event, choose 2 of the 3 calibration specialists and 2 of the 4 field specialists:

\binom{3}{2}\binom{4}{2}=3\cdot 6=18.

Every such committee already satisfies the condition, so the conditional probability is

\frac{18}{34}=\frac{9}{17}\approx 0.5294.

Check. The 18 target committees are part of the 34 restricted committees, and $18/34$ is between 0 and 1. A common mistake is to use 35 as the denominator, giving $18/35\approx 0.5143$. That is the unconditional probability, not the probability given at least one calibration specialist.

Worked Example: Arrangements With Nonadjacent Objects

Seven distinct display units are placed in a row, with every ordering equally likely. Two designated units, $A$ and $B$, are known not to be adjacent. Given that restriction, find the probability that two other designated units, $C$ and $D$, occupy the first two positions, in either order.

Method. Count ordered arrangements. First count the denominator by removing arrangements in which $A$ and $B$ are adjacent. Treating the adjacent pair as one block gives $6!$ arrangements of the block and the other five units; $A$ and $B$ can be ordered inside the block in 2 ways. Thus the number of unrestricted arrangements with $A,B$ adjacent is $2\cdot6!=1{,}440$, and the restricted denominator is

7!-2\cdot6!=5{,}040-1{,}440=3{,}600.

For the numerator, place $C,D$ in the first two positions in either order, giving 2 choices. The remaining five positions contain $A,B$ and three other distinct units. There are $5!$ arrangements of those five units, but $2\cdot4!$ have $A,B$ adjacent. Therefore the numerator is

2\bigl(5!-2\cdot4!\bigr) =2(120-48) =144.

The conditional probability is

\frac{144}{3{,}600}=0.0400.

Check. Of the 144 arrangements in the numerator, none has $A,B$ adjacent, as required. A tempting but incorrect numerator is $2\cdot5!=240$: it includes arrangements where $A,B$ are adjacent, which are excluded by the condition. The restriction must be retained when counting the intersection.

Worked Example: A Code Given Its Number of Even Digits

A three-position code is formed by choosing distinct digits from 1 through 9, with all such codes equally likely. Given that the code contains exactly one even digit, find the probability that its first digit is 2.

Method. The outcome is an ordered code without repeated digits. There are 4 even digits and 5 odd digits. For the restricted denominator, choose the one even digit, choose the two odd digits, and arrange the three chosen digits:

\binom{4}{1}\binom{5}{2}3! =4\cdot10\cdot6 =240.

For the numerator, the first digit must be 2. To have exactly one even digit, the remaining two positions must contain two distinct odd digits. There are 5 choices for the second position and then 4 for the third, giving

5\cdot4=20

codes. Thus

P(\text{first digit is 2}\mid\text{exactly one even digit}) =\frac{20}{240} =\frac{1}{12} \approx 0.0833.

Check. Fixing the first digit at 2 leaves only odd digits for the other positions, so every counted code has exactly one even digit. Using all $9\cdot8\cdot7=504$ codes as the denominator would ignore the given information; the appropriate denominator is 240.

Common Counting Traps

Conditional counting problems often become difficult not because the arithmetic is advanced, but because the counted outcomes do not match the experiment. Check the following points before dividing:

Using the unrestricted denominator. The denominator must count only outcomes satisfying the condition.
Changing whether order matters. A committee is an unordered selection; a code or row arrangement is ordered. Do not use combinations for ordered outcomes or permutations for unordered ones.
Dropping the condition in the numerator. The numerator counts outcomes satisfying both the condition and the target event, not just the target event in isolation.
Counting outcomes that are not equally likely. The ratio of counts works only when the original elementary outcomes are equally likely. For unequal probabilities, sum probabilities instead.
Miscounting overlapping cases. If a condition is split into cases, make them disjoint or account for any overlap so outcomes are not counted twice.

A useful final check is to ask whether every outcome in the numerator is also in the denominator. If not, the restriction has been lost. Also confirm that the denominator is positive and that the resulting probability lies between 0 and 1.

Key takeaway: In an equally likely finite sample space, conditional probability is a ratio of counts: count the outcomes satisfying both the condition and the target event, then divide by the count satisfying the condition alone.

Check Your Understanding

For each problem, identify the restricted denominator before calculating the conditional probability.

A committee of 3 is selected uniformly from 2 actuaries and 5 analysts. Given that it contains at least one actuary, find the probability that it contains exactly one actuary. (Answer: $4/5$)
Six distinct packages are arranged in a row. Given that packages $P$ and $Q$ are not adjacent, find the probability that $R$ and $S$ occupy the first two positions, in either order. (Answer: $1/20$)
A group of 4 is selected uniformly from 5 red and 4 blue tokens. Given that it contains at least one blue token, find the probability that it contains exactly 2 red tokens and includes a specified red token. (Answer: $24/121$)
A four-digit code uses distinct digits from 0 through 8, and its first digit cannot be 0. Given that the code contains exactly 2 even digits, find the probability that its first digit is 2. (Answer: $1/9$)
A group of 5 is selected uniformly from 4 marked and 6 unmarked objects. Given that it contains at least 2 marked objects, find the probability that it contains two specified marked objects. (Answer: $28/93$)

What You'll Learn