Question

A box contains 2 black, 4 white and 3 red balls. One ball is drawn at random from the box, and kept aside. From the remaining balls in the box, another ball is drawn at random and kept besides the first. This process is repeated till all the balls are drawn from the box. The probability that the balls drawn are in the sequence of 2 black, 4 white and 3 red is $\dfrac{1}{140k}$ . Find the value of $k$?

Answer 1

For answering this question we will find the individual conditional probability of picking a ball in the sequence from the box without any replacement of the balls into the box. After that we will find its product which will be equal to the required probability.

Complete step by step answer:
Let us also assume ${{p}_{1}},{{p}_{2}},..........,{{p}_{9}},$ be the probabilities of drawing a black ball, a black ball, a white ball, a white ball, a white ball, a white ball, a red ball, a red ball and a red ball respectively in this order from the box without any replacement of the balls into the box.
Let us also assume the required probability as $p$ then it can be given as ${{p}_{1}}{{p}_{2}}{{p}_{3}}{{p}_{4}}{{p}_{5}}{{p}_{6}}{{p}_{7}}{{p}_{8}}{{p}_{9}}$ .
We will have the value of ${{p}_{1}}$ as $\dfrac{^{2}{{C}_{1}}}{^{9}{{C}_{1}}}$ because we have 2 black balls and total 9 balls in the box.
Similarly for the following values ${{p}_{2}}$ as $\dfrac{^{1}{{C}_{1}}}{^{8}{{C}_{1}}}$ because we have 1 black ball and total 8 balls in the box.
The value of ${{p}_{3}}$ will be $\dfrac{^{4}{{C}_{1}}}{^{7}{{C}_{1}}}$ because we have 4 white balls and total 7 balls in the box.
The value of ${{p}_{4}}$ will be $\dfrac{^{3}{{C}_{1}}}{^{6}{{C}_{1}}}$ because we have 3 white balls and total 6 balls in the box.
The value of ${{p}_{5}}$ will be $\dfrac{^{2}{{C}_{1}}}{^{5}{{C}_{1}}}$ because we have 2 white balls and total 5 balls in the box.
The value of ${{p}_{6}}$ will be $\dfrac{^{1}{{C}_{1}}}{^{4}{{C}_{1}}}$ because we have 1 white ball and total 4 balls in the box.
The value of ${{p}_{7}}$ will be $\dfrac{^{3}{{C}_{1}}}{^{3}{{C}_{1}}}$ because we have 3 red balls and total 3 balls in the box.
The value of ${{p}_{8}}$ will be $\dfrac{^{2}{{C}_{1}}}{^{2}{{C}_{1}}}$ because we have 2 red balls and total 2 balls in the box.
The value of ${{p}_{9}}$ will be $\dfrac{^{1}{{C}_{1}}}{^{1}{{C}_{1}}}$ because we have 1 red ball and total 1 ball in the box.
So the required probability will be $p=\dfrac{^{2}{{C}_{1}}}{^{9}{{C}_{1}}}\dfrac{^{1}{{C}_{1}}}{^{8}{{C}_{1}}}\dfrac{^{4}{{C}_{1}}}{^{7}{{C}_{1}}}\dfrac{^{3}{{C}_{1}}}{^{6}{{C}_{1}}}\dfrac{^{2}{{C}_{1}}}{^{5}{{C}_{1}}}\dfrac{^{1}{{C}_{1}}}{^{4}{{C}_{1}}}\dfrac{^{3}{{C}_{1}}}{^{3}{{C}_{1}}}\dfrac{^{2}{{C}_{1}}}{^{2}{{C}_{1}}}\dfrac{^{1}{{C}_{1}}}{^{1}{{C}_{1}}}$ .
After simplifying this using $^{n}{{C}_{1}}=n$ we will have $p=\dfrac{2}{9}\times \dfrac{1}{8}\times \dfrac{4}{7}\times \dfrac{3}{6}\times \dfrac{2}{5}\times \dfrac{1}{4}\times \dfrac{3}{3}\times \dfrac{2}{2}\times \dfrac{1}{1}$.
By further simplifying this we will have $p=\dfrac{2\times 4\times 3\times 2\times 3\times 2}{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2}$ .
After performing the simplifications we will have $p=\dfrac{1}{9\times 7\times 5\times 4}$ .
After multiplying them we will have $p=\dfrac{1}{9\times 140}$ .

By comparing this with $\dfrac{1}{140k}$ which is given as the required probability in the question we will have $k=9$.

Note: While performing this type of question we should be clear that there is no replacement of the balls in the box. If we consider as there is a replacement then the values of the individual probabilities will be $\dfrac{^{2}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{2}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{4}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{4}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{4}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{4}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{3}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{3}{{C}_{1}}}{^{9}{{C}_{1}}},\dfrac{^{3}{{C}_{1}}}{^{9}{{C}_{1}}}$ respectively. By using them and calculating the required probability we will have a complete mess.