In a box, there are $3$ blue, $4$ red, and $5$ green balls. What is the probability of picking up four balls such that there are $2$ red balls and $1$ ball each of blue and green?

\frac{1}{11}

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\frac{2}{11}

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\frac{3}{11}

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\frac{4}{11}

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Solution

The correct option is B $\frac{2}{11}$

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In a box, there are 3 blue, 4 red, and 5 green balls. What is the probability of picking up four balls such that there are 2 red balls and 1 ball each of blue and green?

The correct option is B 211

In a box, there are $3$ blue, $4$ red, and $5$ green balls. What is the probability of picking up four balls such that there are $2$ red balls and $1$ ball each of blue and green?

The correct option is B $\frac{2}{11}$