Question

12 balls are distributed among three boxes. The probability that the first box contain 3 balls is (a) \((110 / 9)(2 / 3)^{10}\) (b) \((9 / 110)(2 / 3)^{10}\) (c) \(\left[\left({ }^{12} 3\right) / 12^{3}\right] \cdot 2^{9}\) (d) \(\left[\left(\begin{array}{ll}12 & 3\end{array}\right) / 3^{12}\right]\)

Short answer

The probability that the first box contains 3 balls is \(\frac{10}{91}\). This does not match any of the given options, indicating a possible error in the question or the options.

Step-by-step solution

Calculate the total number of distributions

First, we must determine the total number of possible ball distributions among the three boxes. To do this, we can use the formula for distributing n balls into r boxes: \[\left(\begin{array}{ll}n+r-1 & r-1\end{array}\right)\] In our case, we have 12 balls (n=12) and 3 boxes (r=3): \[Total\,Distributions=\left(\begin{array}{ll}12+3-1 & 3-1\end{array}\right)=\left(\begin{array}{ll}14 & 2\end{array}\right)\] Calculate \(\left(\begin{array}{ll}14 & 2\end{array}\right)\): \[Total\,Distributions=\frac{14!}{2!12!}=91\]

Calculate the number of favorable distributions

Now, we must determine the number of distributions where the first box contains 3 balls. Since the first box must have 3 balls, we only need to distribute the remaining 9 balls between the other two boxes. This is a similar problem to step 1, with 9 balls and 2 boxes: \[Favorable\,Distributions=\left(\begin{array}{ll}9+2-1 & 2-1\end{array}\right)=\left(\begin{array}{ll}10 & 1\end{array}\right)\] Calculate \(\left(\begin{array}{ll}10 & 1\end{array}\right)\): \[Favorable\, Distributions=\frac{10!}{1!9!}=10\]

Calculate the probability

Now that we have the total number of distributions and the number of favorable distributions, we can calculate the probability of the first box containing 3 balls: \[Probability=\frac{Favorable\,Distributions}{Total\,Distributions}=\frac{10}{91}\]

Compare the probability with the given options

Finally, compare the calculated probability with the given options: Option (a): \(\frac{110}{9}(2 / 3)^{10}\) Option (b): \(\frac{9}{110}(2 / 3)^{10}\) Option (c): \(\left[\left(\begin{array}{ll}12 & 3\end{array}\right) / 12^{3}\right] \cdot 2^{9}\) Option (d): \(\left[\left(\begin{array}{ll}12 & 3\end{array}\right) / 3^{12}\right]\) None of the given options match our calculated probability \(\frac{10}{91}\). It is possible that there is an error in the question or the options, and we would recommend checking the exercise for possible errors.

Key concepts

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of objects. It helps solve problems related to distributing items into groups or choosing items from a set. In this exercise, we are trying to distribute 12 balls into three boxes. To find the total ways to distribute these balls, we use combinatorial methods.
This approach involves understanding the different ways to arrange items based on constraints. For ball and box problems, combinatorics helps us reason through how items can be split among different groups or places.
Mastering combinatorics involves learning formulas and concepts like permutations and combinations, which allow us to count different arrangements and selections effectively.

Binomial Coefficients

Binomial coefficients are crucial in calculating combinations. They represent the number of ways to choose a set of items from a larger pool, ignoring order. This is mathematically expressed as \( \binom{n}{k} \), which means "n choose k." In our scenario, this formula helps calculate how many ways we can arrange a certain number of objects (balls) into groups (boxes).
For example, the formula \( \binom{14}{2} \) was used to find the total distributions in our problem. This coefficient is part of the binomial theorem, which provides a compact way to handle large combinations efficiently.

Ball and Box Problems

Ball and box problems involve distributing indistinguishable items (balls) into distinct groups (boxes). These types of problems often appear in probability theory and combinatorics.
The formula \( \binom{n+r-1}{r-1} \) is used for situations where items are indistinguishable, and we distinguish between boxes. In our task, we tackled the challenge of distributing 12 balls into 3 boxes while ensuring a particular condition (3 balls in the first box).
This approach involves breaking down the total set of distributions into favorable and total distributions, allowing us to find specific probabilities based on constraints.

Probability Calculation

Probability calculation involves determining the likelihood of an event occurring. In this exercise, we needed to find the probability that one specific box contains 3 balls. The probability is a ratio of favorable outcomes to the total possible outcomes.
Given our calculations, the probability was \( \frac{10}{91} \), found by dividing the number of favorable distributions (ways one box has 3 balls) by the total distributions (all possible ways balls can be distributed).
Understanding probability involves assessing every possible outcome to gain accurate insights into the likelihood of desired events. Mastery in this area aids in making predictions and informed decisions based on possible results.