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# A box contain 10 red marbles,20 blue marbles and 30 green marbles.5 marbles are drawn from the box.What is the probability that atleast one will be green?

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

Toolbox:
• $nC_r=\large\frac{n!}{r!(n-r)!}$
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.