Question

An urn contains marbles of four colours: red, white, blue and green. When four marbles are drawn without replacement, the following events are equally likely:
$\left(1\right)$ the selection of four red marbles;
$\left(2\right)$ the selection of one white and three red marbles;
$\left(3\right)$ the selection of one white, one blue and two red marbles;
$\left(4\right)$ the selection of one marble of each colour.
The smallest total number of marbles satisfying the given condition is.

• A. $19$
• B. $21$
• C. $46$
• D. $69$

Answer 1

Answer: (B)

$21$

Let number of red, white, blue, green balls be $r,w,b,g$ respectively and $r+w+b+g=n$
Given $\frac{{}^r{C}_4}{{}^n{C}_4}=\frac{{}^w{C}_1{.}^r{C}_3}{{}^n{C}_4}=\frac{{}^w{C}_1{.}^b{C}_1{.}^r{C}_2}{{}^n{C}_4}=\frac{{}^r{C}_1{.}^w{C}_1{.}^b{C}_1.g{C}_1}{{}^n{C}_4}$
$\Rightarrow \frac{r(r-1)(r-2)(r-3)}{24}=\frac{wr(r-1)(r-2)}{6}=\frac{wbr(r-1)}{2}=r.w.b.g$
$\Rightarrow r-3=4w,r-2=3b,r-1=2g$
$\Rightarrow r=4w+3,r=3b+2,r=2g+1$
Now $LCM$ of $(4,3,2)=12$
$\therefore r_{min}=11\Rightarrow w_{min}=2\Rightarrow b_{min}=3\Rightarrow g_{min}=5$
$\therefore (r+w+b+g{)}_{min.}=21$.