A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

Engineering

Mathematics

Conditional Probability

Question

$\frac{2}{9}$

$\frac{1}{14}$

$\frac{2}{63}$

$\frac{2}{14}$

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Total Blue Pens: 4

Total Red Pens: 2

Total Black Pens: 3

Total Pens: 4 + 2 + 3 = 9

Probability of drawing 2 blue pens $= \frac{^{4} C_{2}}{^{9} C_{2}} = \frac{4 \times 3}{9 \times 8} = \frac{1}{6}$

After this, the pens are not replaced, which reduces the number of pens in the pack to 7.

So, the probability of drawing 1 black pen from a pack of 7 pens would be $\frac{^{3} C_{1}}{^{7} C_{1}} = \frac{3}{7}$

Probability of drawing 2 blue pens and 1 black pen $= \frac{1}{6} \times \frac{3}{7} = \frac{1}{14}$ .

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