$\begin{array}{1 1}(A)\;1-\large\frac{30C_5}{60C_5}\\(B)\;1-\large\frac{40C_5}{80C_5}\\(C)\;1-\large\frac{20C_5}{60C_5}\\(D)\;1-\large\frac{50C_5}{60C_5}\end{array} $

- $nC_r=\large\frac{n!}{r!(n-r)!}$

Step 1:

Given :

10 Red marbles

20 Blue marbles

30 Green marbles

5 marbles are drawn at random

$\therefore$ Total number of ways selecting 5 marbles out of 60 marbles =$60C_5$

Step 2:

Atleast one will be green

Probability of atleast one will be green =1-Probability of no green

Probability of no green=$\large\frac{30C_5}{60C_5}$

$\Rightarrow 1-\large\frac{30C_5}{60C_5}$

Hence (A) is the correct answer.

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