$\begin{array}{1 1}(A)\;\large\frac{3}{7}\\(B)\;\large\frac{2}{9}\\(C)\;\large\frac{1}{9}\\(D)\;\text{None of these}\end{array} $

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Step 1:

Let the probability of D selected be $'x'$

$\therefore P(D)=x$

$P(C)=2x$

$P(B)=2x$

$P(A)=4x$

$P(A)+P(B)+P(C)+P(D)=1$

Sum of all probabilities =1

$\therefore 4x+2x+2x+x=1$

$9x=1$

$x=\large\frac{1}{9}$

Step 2:

C will be selected

$\therefore P$(C selected)=$2x$

$\Rightarrow 2\times \large\frac{1}{9}=\frac{2}{9}$

Hence (B) is the correct answer.

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