Question

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is

• A. $\frac{1}{3}$
• B. $\frac{2}{7}$
• C. $\frac{1}{21}$
• D. $\frac{2}{23}$

Answer 1

The correct option is B

$\frac{2}{7}$

Explanation for the correct option:

Step: 1 Calculate the number of ways $3$balls can be selected at random from $9$balls.

The urn contains a total of $9$balls.

Three balls are drawn at random without replacement.

Therefore, different ways $3$ balls can be drawn from $9$ balls is $C_3^9$.

Step: 2 Calculate the number of ways $1$ ball of each color can be selected at random from the balls of their respective color.

Again, the different ways $1$ red ball can be drawn from $3$ red balls is $C_1^3$.

The different ways $1$ blue ball can be drawn from $4$ blue balls is $C_1^4$.

The different ways $1$ green ball can be drawn from $2$ green balls is $C_1^2$.

Step: 3 Calculate the probability that $3$ balls selected at random have different colors.

Now, the probability that $3$ balls selected at random have different colors is

$$\begin{gathered} =\frac{C_1^3\times C_1^4\times C_1^2}{C_3^9} \\ =\frac{3\times 4\times 2}{{\displaystyle \frac{9!}{3!\times 6!}}} \\ =\frac{24}{{\displaystyle \frac{9\times 8\times 7\times 6!}{3\times 2\times 6!}}} \\ =\frac{24}{84} \\ =\frac{2}{7} \end{gathered}$$

Therefore, Option(B) is the correct answer.