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The scenario of one of my previous questions is described below:

Suppose we have $12$ identical balls. Think of the balls as having distinct ID numbers, $1-12$. We have $3$ identical boxes, each of which can contain exactly $4$ balls. One after the other the balls are thrown at random into one of the three boxes.

Suppose Mr. A will buy one ball from each box, so he will buy a total of three balls from the three boxes. Habit of Mr. A is to buy that ball from the box which ID is smallest of all other balls in that box.

For example, the randomly thrown balls in a box have ID $4$, $8$, $9$, and $12$. So Mr. A will buy the ball with ID $4$ from that box.

Then Mr. A goes to another box and here the balls have ID $1$, $3$, $6$, and $11$. He will buy the ball with ID $1$ from that box.

At last, Mr. A goes to the last box and here the balls have ID $2$, $5$, $7$, and $10$. He will buy the ball with ID $2$ from that box.

The probability that Mr. A will buy the ball with ID $i$, where $i=1,2,\ldots,12$ is $$\frac{\binom{12-i}3}{\binom{11}3}.$$

Now my question is

If the same experiment is repeated $10$ times, then what is the probability that Mr. A will buy the ball with ID $i$, where $i=1,2,\ldots,12$?

Simply multiplying $10$ with $\frac{\binom{12-i}3}{\binom{11}3}$ makes the probability greater than $1$.

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The probability that Mr. A do not buy the ball with ID i in $10$ times is

$$P(X_{10}=0)=\left(1-\frac{\binom{12-i}{3}}{\binom{11}{3}}\right)^{10} $$

Now the converse probability can be used. The probability that Mr. A do buy at least one ball with ID i in $10$ times is

$$P(X_{10}\geq 1)=1-P(X_{10}=0)=1-\left(1-\frac{\binom{12-i}{3}}{\binom{11}{3}}\right)^{10}$$

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    Could you please tell what will be $P(X_{10}=1)$? – 2017-01-29
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    @ABC $P(X_{10}=1)$ is the probability that Mr. A buy the ball with ID i once in 10 times. – 2017-01-29
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    Yes. I understand. But can't guess the formula. Just want to know for curiosity. – 2017-01-29
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    @ABC $X_{10}$ is binomial distributed as $X\sim Bin(10,p)$, where p is is the fraction with the binomial coeffiecients. It can be written in a different way: $P(X_{10}=0)=\binom{10}{0}\cdot p^0\cdot (1-p)^{10}$. The first two factors are both just 1. – 2017-01-29
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    @ABC I´ve finished my previous comment. – 2017-01-29
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    So $P(X_{10}=1)=\binom{10}1(\frac{\binom{12-i}3}{\binom{11}3})(1-\frac{\binom{12-i}3}{\binom{11}3})^9 $. – 2017-01-29
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    @ABC That´s correct. – 2017-01-29