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A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at

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Bunuel
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A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at [#permalink] A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at   05 Apr 2023, 01:30
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A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen?

(A) \(\frac{16}{243}\)

(B) \(\frac{12}{81}\)

(C) \(\frac{16}{81}\)

(D) \(\frac{3}{9}\)

(E) \(\frac{4}{9}\)



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Re: A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at [#permalink] A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at   05 Apr 2023, 03:44
Since, a pen is drawn and then replaced, the second draw's result is independent from the first draw's result.
Thus, the required probability (2Blue and 1 Black pen) = \(\frac{4}{9}*\frac{4}{9}*\frac{3}{9}\) = \(\frac{16}{243}\)
Answer A
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Re: A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at [#permalink] A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at   05 Apr 2023, 07:00
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Bunuel wrote:
A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen?

(A) \(\frac{16}{243}\)

(B) \(\frac{12}{81}\)

(C) \(\frac{16}{81}\)

(D) \(\frac{3}{9}\)

(E) \(\frac{4}{9}\)




We need to get the following:

Blue-Blue-Black (not necessarily in that order)

So, we have:

4/9 x 4/9 x 3/9

4/9 x 4/9 x 1/3

Since Blue-Blue-Black can be arranged in 3!/2! = 3 ways, the final probability is:

4/9 x 4/9 x 1/3 x 3 = 16/81

Answer: C
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Re: A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at [#permalink]
05 Apr 2023, 07:00
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Bunuel
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