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Chapter 18: Problem 788

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

Expert verified
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

01

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 \(\left(\begin{array}{ll}14 & 2\end{array}\right)\): \[Total\,Distributions=\frac{14!}{2!12!}=91\]
02

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 \(\left(\begin{array}{ll}10 & 1\end{array}\right)\): \[Favorable\, Distributions=\frac{10!}{1!9!}=10\]
03

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: \[Probability=\frac{Favorable\,Distributions}{Total\,Distributions}=\frac{10}{91}\]
04

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.

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Most popular questions from this chapter

A three digit number which is a multiple of 11 is chosen at random. The probability the number so chosen is also a multiple of 9 is (a) \((1 / 9)\) (b) \((2 / 9)\) (c) \((1 / 100)\) (d) \((9 / 100)\)

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