Question

A lucky draw machine is filled with $10$ colored balls as $5$ red, $3$ green, and $2$ blue. Machine selects $2$ balls one by one withouth replacement. What is the probability that no two balls are of the same color?

• A. $\frac{31}{45}$
• B. $\frac{14}{45}$
• C. $\frac{13}{45}$
• D. $\frac{32}{45}$

Answer 1

The correct option is A $\frac{31}{45}$

Given that:
Total number of balls $=10$
Number of red balls $=5$
Number of green balls $=3$
Number of blue balls $=2$

$\Rightarrow$Total number of ways of selecting $2$ balls out of $10$
$={}^{10}{\text{C}}_2$
$=\frac{10\times 9}{2}=45$
Total number of ways of selecting $2$ red balls out of $5$
$={}^5{\text{C}}_2$
$=\frac{5\times 4}{2}=10$
Total number of ways of selecting $2$ green balls out of $3$
$={}^3{\text{C}}_2$
$=\frac{3\times 2}{2}=3$
Total number of ways of selecting $2$ blue balls out of $2$
$={}^2{\text{C}}_2=1$

Now, find the probability of secting $2$ balls of same color

Probability of selecting $2$ red balls
$=\frac{10}{45}$
$=\frac{2}{9}$
Probability of selecting $2$ green balls
$=\frac{3}{45}$
$=\frac{1}{15}$
Probability of selecting $2$ blue balls
$=\frac{1}{45}$

$\Rightarrow$ So, probablity of selecting two balls of same color
$=\frac{2}{9}+\frac{1}{15}+\frac{1}{45}$
$=\frac{10+3+1}{45}$
$=\frac{14}{45}$

Hence,
Probability of selecting two balls not of same color
$=1-$ Probability of selecting two balls of same color
$=1-\frac{14}{45}$
$=\frac{31}{45}$