Question

A bag contains 3R and 3G balls and a person draws out 3 at random. He then drops 3 blue balls into the bag and again draws out 3 at random. The chance that the 3 later balls being all of different colors is ?
A.$15\% $
B.$20\% $
C.$27\% $
D.$40\% $

Answer 1

Hint- To solve this question, we will use the basic definition of probability which states that it is the ratio of favorable outcomes to the total number of outcomes. First we will define the event and then using combination for selection we will proceed further.

Complete step-by-step answer:
Given that
Red balls = 3
Green balls = 3
Blue balls = 3
The man draws 3 balls at random from the bag which contains 3R and 3G balls, the possible cases are
a)1R+2G is drawn, 2R and 1G balls left in the bag
b)2R+1G is drawn, 1R and 2G balls left in the bag
c)3R is drawn, 3G balls left in the bag
d)3G is drawn, 3R balls left in the bag


Now, the man drops 3 blue balls in the bag, total numbers of balls in the bag is 6 again
We need to find the probability that the 3 later balls of different colors
Therefore possible numbers of cases are
a)2R+1G+3B
b)1R+2G+3B

Probability of withdrawing 2R and 1G ball
$
   = \dfrac{{^3{C_2}{ \times ^3}{C_1}}}{{^6{C_3}}} \\
   = \dfrac{9}{{20}} \\
$
Probability of withdrawing 1R,1G, 1B from case 1
$
   = \dfrac{{^2{C_1}{ \times ^1}{C_1}{ \times ^3}{C_1}}}{{^6{C_3}}} \\
   = \dfrac{6}{{20}} \\
$
 For second case
Probability of withdrawing 1R and 2G ball
\[
   = \dfrac{{^3{C_1}{ \times ^3}{C_2}}}{{^6{C_3}}} \\
   = \dfrac{9}{{20}} \\
\]
Probability of withdrawing 1R,1G, 1B from case 1
$
   = \dfrac{{^2{C_1}{ \times ^1}{C_1}{ \times ^3}{C_1}}}{{^6{C_3}}} \\
   = \dfrac{6}{{20}} \\
$

So, chance =
\[
   = \dfrac{9}{{20}} \times \dfrac{6}{{20}} \times 100 \\
   = 27\% \\
\]
Hence, the chance of getting drawing the three balls at later stage is \[27\% \]

Note- In order to solve these types of questions, remember the basic definition of probability and first create a sample space and define events. In the above question the event is already defined that is of drawing the ball from the bag. Similarly in other questions different events can be defined like getting only heads etc. The process to solve the question remains the same.