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A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. Then, the probability that none of the balls drawn is blue is
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Verified Answer
The correct answer is:
$10 / 21$
Given bag contains 2 red, 3 green 2 blue balls
$$
\begin{aligned}
& \mathrm{P} \text { (none ball drawn is blue) }=\frac{{ }^2 \mathrm{c}_2+{ }^3 \mathrm{c}_2+{ }^2 \mathrm{c}_1+{ }^3 \mathrm{c}_1}{{ }^7 \mathrm{c}_2} \\
& =\frac{1+3+2 \times 3}{\frac{7 \times 6}{2}}=\frac{4+6}{7 \times 3}=\frac{10}{21}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{P} \text { (none ball drawn is blue) }=\frac{{ }^2 \mathrm{c}_2+{ }^3 \mathrm{c}_2+{ }^2 \mathrm{c}_1+{ }^3 \mathrm{c}_1}{{ }^7 \mathrm{c}_2} \\
& =\frac{1+3+2 \times 3}{\frac{7 \times 6}{2}}=\frac{4+6}{7 \times 3}=\frac{10}{21}
\end{aligned}
$$
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