Question

A bag contains $3$ red, $4$ white and $5$ black balls. Three balls are drawn at random. The probability of being their different colors is

• A. $\frac{3}{11}$
• B. $\frac{2}{11}$
• C. $\frac{8}{11}$
• D. none of these

Answer 1

The correct option is A

$\frac{3}{11}$

Explanation for correct option:

Total number of balls in a bag are $3+4+5=12$

Three balls drawn at random is ${}^{12}C_3$

When The all three ball are drawn different ${}^3{C}_1\times {}^{4}C_1\times {}^5{C}_1$

Now probability that three ball drawn from bag random and different colors is $\frac{{}^{3}C_1\times {}^{4}C_1\times {}^{5}C_1}{{}^{12}C_3}$

By using formula ${}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{\times }\mathbf{r}\mathbf{!}}$

$$\begin{gathered} \frac{{}^{3}C_1\times {}^{4}C_1\times {}^{5}C_1}{{}^{12}C_3}=\left(\frac{{\displaystyle \frac{3!}{1!\times 2!}}\times {\displaystyle \frac{4!}{1!\times 3!}}\times {\displaystyle \frac{5!}{1!\times 4!}}}{{\displaystyle \frac{12!}{9!\times 3!}}}\right) \\ =\frac{3\times 4\times 5\times 6}{12\times 11\times 10} \\ =\frac{3}{11} \end{gathered}$$

Therefore, The probability that the three balls drawn from bag of being their different colors is $\frac{3}{11}$.

Thus correct answer is option $\left(A\right)$.