Question

12 balls are distributed among three boxes and find the probability that the first box will contain 3 balls.

Answer 1

- Hint:-In this question first we have to find the number of ways in which we can place all 12 balls. Then we have to find a number of ways in which we can place three balls in the first box and the remaining 9 balls in the other two boxes. After finding these apply the formula of probability.

Complete step-by-step solution -

Given: 12 balls are distributed among three boxes and we have to find the probability that the first box will contain 3 balls.
Now, each ball can be put into any of three boxes.
Therefore, total number of ways in which 12 balls can be put into three boxes = \[{3^{12}}\]
Again, out of 12 given balls , three balls can be chosen in $_{}^{12}{C_3}$.
Now remaining balls are 12-3=9.
These remaining 9 balls we have to put in two other boxes.
Therefore, total number of ways in which these remaining 9 balls can be put into two boxes=${2^9}$
We know probability of finding favourable outcome is given by
\[ \Rightarrow P(E) = \dfrac{{{\text{Number of favourable outcome}}}}{{{\text{Number of all possible outcomes}}}}\].
Now, let the probability that first box will contain 3 balls be$P(E)$$ \Rightarrow P(E) = \dfrac{{{\text{Number of ways in which first box will contain 3 balls and remaining 9 balls will contain by other 2 boxes}}}}{{{\text{Total number of ways in which 12 balls can be put into three boxes}}}}$
$ \Rightarrow P(E) = \dfrac{{_{}^{12}{C_3} \times {2^9}}}{{{3^{12}}}}$
On solving above equation we get
$
   \Rightarrow P(E) = \dfrac{{12 \times 11 \times 10 \times {2^9}}}{{3! \times {3^{12}}}} \\
   \Rightarrow P(E) = \dfrac{{12 \times 11 \times 10 \times {2^9}}}{{6 \times {3^{12}}}}{\text{ \{ 3! = 6\} }} \\
   \Rightarrow P(E) = \dfrac{{2 \times 11 \times 10 \times {2^9}}}{{{3^{12}}}} \\
$
$
 \Rightarrow P(E) = \dfrac{{110 \times {2^{10}}}}{{9 \times {3^{10}}}} \\
   \Rightarrow P(E) = \dfrac{{110}}{9}{\left( {\dfrac{2}{3}} \right)^{10}} \\
 $
Hence, the probability that first box will contain 3 balls is $\dfrac{{110}}{9}{\left( {\dfrac{2}{3}} \right)^{10}}$.

Note:- Whenever you get this type of question the key concept to solve this is to learn the different concepts of permutation and combination and probability like distribution of n things among r distinct objects is ${{\text{r}}^{\text{n}}}$. One more thing to be remember is that probability of finding favourable outcome P(E)$\left\{ {P(E) = \dfrac{{{\text{Number of favourable outcome}}}}{{{\text{Number of all possible outcomes}}}}} \right\}$.