In a small town there are $6$ repairmen and one day $10$ households call for TV repairmen

Question

Question

What is the probability that precisely $2$ of the $6$ repairmen will be called?

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The actual question is more generic but I've used a specific example here in the hope that I can generalise it. However, I think I've made a mistake somewhere because when I tried to simulate the outcomes I got a simulated probability of around 86% yet my theoretical solution yields a value closer to 47%.

I'd appreciate it if someone could point out where I've gone wrong in my solution (below).

My (erroneous) attempt

I thought of the repairmen as faces on a 6-sided die and the number of households, 10, as a series of die rolls. Therefore, I believe that by figuring out the probability of precisely 2 faces of the die appearing in a trial of 10 rolls I am solving an equivalent problem.

First, I defined the event: $$F_j := \{\text{Face $j$ appears 2 times in a series of 10 rolls}\}, \space \text{where} \space j \in \{1,...,6\}$$

So, supposing that $j = 5$ for instance, an example of an outcome in $F_{j = 5}$ might be,

$$\{5, 1, 2, 6, 5, 6, 6, 4, 4, 3\}$$

So for a given face $j$ (which we've set to 5 in this case), there are $5^8$ choices for the remaining 8 rolls. In addition, face $j$ could appear in any of ${10 \choose 2}$ positions. It follows that,

$$|F_j| = {10 \choose 2} \cdot 5^8$$

for a fixed choice of $j$. Further to this, there are ${6 \choose 1}$ ways of choosing $j$. Hence,

$$\left| \bigcup_{j = 1}^{6} F_j \right| \approx {6 \choose 1} {10 \choose 2} 5^8$$

I am aware that these aren't exactly equal because $F_j \cap F_k \neq \emptyset$ (i.e. it is possible for faces $j$ and $k$ to both appear 2 times in 10 rolls). So I then used the inclusion-exclusion principle to come up with the following formula,

$$\left| \bigcup_{j = 1}^{6} F_j \right| = {6 \choose 1} {10 \choose 2} 1^2 5^8 - {6 \choose 2} {10 \choose 4} 2^4 4^6 + ... + {6 \choose 5} {10 \choose 10} 5^{10}1^0$$

However, when I divide this by $6^{10}$ I get an answer of 47% which, as mentioned, doesn't appear to tie up with a quick empirical check I performed in Excel (which could also be wrong!). Where have I gone wrong?

Answer 1

The problem here is that your original translation of the problem was

precisely two faces of the die appear in ten rolls

which is correct. However, you then solved the different problem

some face of the die appears precisely twice in ten rolls.

Going back to the original problem, how many different possibilities are there for the two faces which appear? For each such pair, what is the probability that precisely those two faces appear in the ten rolls (i.e. each of them appears at least once, and no others appear)?