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

Question

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

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

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

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

(A) \frac{16}{243}

(B) \frac{16}{81}

(C) \frac{4}{9}

(D) \frac{12}{81}

(E) \frac{3}{9}

chetan2u · Answer

Total ways = 9*9*9=729

Probability of picking blue = 1/4 and of black = 1/3
So probability of drawing 2 blue pens and 1 black pen =  (\frac{4}{9})^2*\frac{3}{9} , but 2 blue and 1 black can be picked up in  \frac{3!}{2!}=3  ways
Probability =  (\frac{4}{9})^2*\frac{3}{9}*3=\frac{4*4*3*3}{9*9*9}=\frac{16}{81}

B
OA is wrong, as the answer would be 16/243 when we are looking at a certain pattern say black, blue , black. And we can have three such orders , so 3*16/243

karan12345 · Answer

Yes, but I have updated the question that it is in a certain pattern

chetan2u · Answer

Please dont change question after it has been answered as per original. Creates confusion in other members seeking an answer

bumpbot · Answer

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

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