Question

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?
(a) \[\dfrac{2}{9}\]
(b) \[\dfrac{1}{14}\]
(c) \[\dfrac{2}{63}\]
(d) \[\dfrac{2}{14}\]

Answer 1

Hint:Solve this question by considering the 3 steps. The first is to allow the first blue pen, second is to draw a second blue pen and third is to draw a black pen and all these without replacement. So, use the probability of drawing a pen \[=\dfrac{\text{Remaining pens of a particular color}}{\text{Remaining total pens}}\]. Multiply all three steps to get the required answer.

Complete step-by-step answer:
We are given that 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. We have to find the probability of drawing 2 blue pens and 1 black pen. We have a total of 4 + 2 + 3 = 9 pens out of which 4 are blue, 2 are red and 3 are black pens.
We have to draw 2 blue pens and 1 black pen without a replacement. We will do it in a step by step manner.
1. Here, we have 9 pens comprising 4 blue pens, 2 red pens, and 3 black pens. So, we get,
Probability of drawing first blue pen \[=\dfrac{\text{Total blue pens}}{\text{Total pens}}\]
\[=\dfrac{4}{9}....\left( i \right)\]
2. As the blue pen is not replaced, so now we are remaining with 8 pens comprising 3 blue pens, 2 red pens, and 3 black pens. So, we get,
Probability of drawing a second blue pen \[=\dfrac{\text{Remaining blue pens}}{\text{Remaining total pens}}\]
\[=\dfrac{3}{8}....\left( ii \right)\]
3. As the blue pens are not replaced. So, now we are remaining with 7 pens comprising 2 blue pens, 2 red pens, and 3 black pens. So, we get,
Probability of drawing a black pen \[=\dfrac{\text{Remaining black pens}}{\text{Remaining total pens}}\]
\[=\dfrac{3}{7}....\left( iii \right)\]
So, now we get the probability of drawing 2 blue pens and 1 black pen = (Probability of drawing first blue pen) \[\times \] (Probability of drawing second blue pen) \[\times \] (Probability of drawing a black pen)
By substituting the values of the terms in the RHS from equation (i), (ii) and (iii) accordingly, we get,
Probability of drawing 2 blue pens and 1 black pen \[=\dfrac{4}{9}\times \dfrac{3}{8}\times \dfrac{3}{7}=\dfrac{1}{14}\]
So, we get the required value of the probability as \[\dfrac{1}{14}\].
So, option (b) is the right answer.

Note: Students must note that in the questions of probability without replacement, the sequence in which the items are drawn matter a lot unlike the case when we do it with replacement. For example, in the case of without replacement, in the above question, drawing a black pen after a blue pen and before a blue pen would have different probabilities, so this must be taken care of. The sequence of items drawn must be properly understood and always make the list of each remaining item and total items after each item is drawn to easily solve the question.