How to Find the Probability of Selecting One Student from Each of Three Cities

The answer depends on how the students are selected. The usual interpretation is that all students from the three cities are placed in one pool, then three students are chosen at random without replacement. Under that assumption, you can count the groups that contain one student from each city and divide by the number of possible groups of three.

If the selection is instead made separately—one draw from each city—then the desired outcome is guaranteed, provided each city has at least one student to choose from.

Formula for one combined pool

Let the three cities contribute $a$, $b$, and $c$ students, respectively. The total number of students is

$$N=a+b+c.$$

When choosing three students without regard to order, the total number of possible groups is

$$\binom{N}{3}.$$

To form a group with exactly one student from each city, choose one of the $a$ students in the first city, one of the $b$ in the second, and one of the $c$ in the third. The number of favorable groups is therefore $abc$. Thus,

$$P(\text{one from each city})=\frac{abc}{\binom{a+b+c}{3}}.$$

This formula assumes $N\geq 3$, that every group of three students is equally likely to be selected, and that nobody can be selected more than once. If one of the cities has no students in the pool, the probability is zero.

Worked example

Suppose the pool contains 8 students from City A, 10 from City B, and 12 from City C. Three students are chosen at random without replacement.

There are $8+10+12=30$ students altogether. The number of possible groups is

$$\binom{30}{3}=\frac{30\cdot29\cdot28}{3\cdot2\cdot1}=4060.$$

For a group with one student from each city, there are 8 choices for the City A student, 10 for City B, and 12 for City C. So the number of favorable groups is

$$8\cdot10\cdot12=960.$$

The probability is

$$\frac{960}{4060}=\frac{48}{203}\approx 0.2365.$$

So the chance is about 23.65%.

Why the order does not change the answer

You can also imagine drawing the three students one at a time. There are six possible orders in which the cities could appear: ABC, ACB, BAC, BCA, CAB, and CBA. For example, the probability of drawing A, then B, then C is

$$\frac{8}{30}\cdot\frac{10}{29}\cdot\frac{12}{28}.$$

Each of the six city orders has the same product of city counts in its numerator, so adding their probabilities gives

$$6\cdot\frac{8\cdot10\cdot12}{30\cdot29\cdot28} =\frac{48}{203}.$$

The factor of six is needed in the sequential calculation because the order can vary. The combination formula already treats all orders as the same group, so it does not need that extra factor.

If you draw one student from each city

If the procedure says to select one student from City A, one from City B, and one from City C, then the event is built into the sampling method. Its probability is 1, assuming each city has at least one eligible student. That differs from drawing three students from a combined pool, where the city of origin of each draw is random.

If the selection probabilities differ—for example, some students are more likely to be included than others—the counts alone are not enough to determine the probability. You would need the sampling rules or the individual selection probabilities.

Common mistakes

• Using the combined-pool formula for separate city-by-city draws. Separate draws already guarantee one from each city; a pooled draw does not.
• Counting only one order. In a sequential calculation, ABC is only one of six possible orders. Include all six, or use combinations directly.
• Using $N^3$ as the number of possible selections. That would describe ordered draws with replacement. Here, selection is without replacement, and groups are unordered, so use $\binom{N}{3}$.
• Mixing up exactly one from each city with at least one from each city. When exactly three students are chosen and there are three cities, these conditions are equivalent. With more students or more cities, they need not be.

If you know the three city counts, substitute them for $a$, $b$, and $c$ in the formula. If the problem specifies a different selection procedure, use that procedure instead of assuming a combined pool.