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 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 $\frac{1}{140k}$. Find the value of $k$.

Answer 1

Let $p_1,p_2,p_3,\cdots ,p_9,$

denote the probabilities of drawing black, black, white, white, white, white, red, red and red respectively in this order without replacement.

Then $p$ = the required probability $=p_1,p_2,p_3,\cdots ,p_9.$ $p_1=\frac{{}^2{C}_1}{{}^9{C}_1}=\frac{2}{9},$

Since one black ball can be drawn out of $2$ in ${}^2{C}_1$ ways and total number of ways is ${}^9{C}_1.$ $p_2=\frac{{}^1{C}_1}{{}^8{C}_1}=\frac{1}{8},$

Since one black ball remains after the first draw.

$p_3=\frac{{}^4{C}_1}{{}^7{C}_1}=\frac{4}{7},$

since in the remaining $7$ balls $4$ are white. Similary,

$p_4=\frac{{}^3{C}_1}{{}^6{C}_1}=\frac{3}{6}=\frac{1}{2},$ $p_5=\frac{{}^2{C}_1}{{}^5{C}_1}=\frac{2}{5},p_6=\frac{1}{4},$

Now the remaining three balls are all red so theat $p_7=p_8=p_9=1.$

Hence $p=\frac{2}{9}.\frac{1}{8}.\frac{4}{7}.\frac{1}{2}.\frac{2}{5}.\frac{1}{4}.=\frac{1}{1260}.$