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

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

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

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

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

(A)  \frac{16}{243}

(B)  \frac{12}{81}

(C)  \frac{16}{81}

(D)  \frac{3}{9}

(E)  \frac{4}{9}

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ScottTargetTestPrep · Answer

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

PSKhore · Answer

Since we are replacing the pens, it will have no effect on the denominator.
But we have to take into account the case, where the 2 blue pens are the same.

(4/9)* (4/9)* (3/9)* (3!/2!)

=  16/81

keets02 · Answer

When they say replace, does it not have an effect on the denominator in the subsequent draws based on what it is replaced with?

Bunuel · Answer

Replacement means that after each draw, the pen picked is put back into the pack, so the total contents of the pack remain unchanged.

That’s why the denominator stays 9 for every draw, each draw is independent and has the same probabilities: 4/9 for blue, 2/9 for red, and 3/9 for black.

If it were  without  replacement, then the denominator would change (first 9, then 8, then 7, etc.), but here replacement keeps it fixed at 9.

findingmyself · Answer

First let's call out all possible combinations in which three pens can be drawn. For this we have to assume from question stem that Only black and blue are been drawn and not red,.

Black Blue Blue
Black Blue Black
Black Black Blue
Black Black Black

Blue Black Blue
Blue Black Black
Blue Blue Black
Blue Blue Blue

From this the outcomes with 2 Blue and 1 Black to be picked.

Black Blue Blue
Blue Black Blue
Blue Blue Black

Now let's find probabilities, since there is replacement, our job is fairly easy

Black Blue Blue=  \frac{3}{9} * \frac{4}{9} * \frac{4}{9} = \frac{48}{729} 
Blue Black Blue= Same as above=  \frac{48}{729} 
Blue Blue Black= Same as above=  \frac{48}{729}

Summing the above three, we will have total probability:  \frac{48*3}{729} = \frac{16}{81}

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

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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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