Question

There is a box which contains 4 red, 3 blue and 1 black identical balls. For each red ball withdrawn (without replacement), you have to put 1 blue and 1 black ball in the bag. For each blue ball withdrawn (without replacement), you have to put 2 blue and 1 red ball. What is the probability that ball withdrawn at third attempt will be black, if first and second balls withdrawn were red and blue respectively?

• A. $\frac{1}{11}$
• B. $\frac{2}{11}$
• C. $\frac{3}{11}$
• D. $\frac{4}{11}$

Answer 1

The correct option is B

$\frac{2}{11}$

Box:$$\begin{array}{ccc}\text{4 Red}& \text{3 Blue}& \text{1 Black}\end{array}$$

Step 1: 1.1: Take out 1 red ball, remaining red balls = 4 - 1 = 3

1.2: Add 1 blue, total blue balls = 3 + 1 = 4

1.3: Add 1 Black, total black balls = 1+ 1 = 2

Box: $$\begin{array}{ccc}\text{3 Red}& \text{4 Blue}& \text{2 Black}\end{array}$$

Step 2: 2.1: Take out 1 blue ball, remaining blue balls = 4 - 1 = 3

2.2: Add 2 blue balls, total blue balls = 3 + 2 = 5

2.3: Add 1 red ball, total red ball = 3 + 1 = 4

Box: $$\begin{array}{ccc}\text{4 Red}& \text{5 Blue}& \text{2 Black}\end{array}$$

P (getting black ball) = $\frac{\text{Total no. of Black balls}}{\text{Total no. of balls}}$= $\frac{2}{5+4+2}$= $\frac{2}{11}$