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 :
(1) the selection of four red marbles;
(2) the selection of one white and three red marbles;
(3) the selection of one white, one blue and two red marbles;
(4) 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

The correct option is B $21$

Let number of red, white, blue, green balls be $R,W,B,G$ respectively and $R+W+B+G=n.$
Given that four marbles are drawn without replacement, and events are equally likely
$\frac{{}^R{C}_4}{{}^n{C}_4}=\frac{{}^W{C}_1{}^R{C}_3}{{}^n{C}_4}=\frac{{}^W{C}_1{}^B{C}_1{}^R{C}_2}{{}^n{C}_4}=\frac{{}^W{C}_1{}^B{C}_1{}^R{C}_1{}^G{C}_1}{{}^n{C}_4}$
$\Rightarrow \frac{R(R-1)(R-2)(R-3)}{24}=\frac{W.R.(R-1)(R-2)}{6}=\frac{W.B.R(R-1)}{2}=W.B.R.G$
From above expression we get,
$R=4W+3=2+3B=1+2G$
Now L.C.M of $(4,3,2)=12$
For $R_{min}=11\Rightarrow W_{min}=2\Rightarrow B_{min}=3\Rightarrow G_{min}=5$
$\therefore (R+W+B+G{)}_{min}=21$