Question

An urn contains $8$ red and $5$ white balls. Three balls are drawn at random. Then, the probability that balls of both colors are drawn, is

• A. $\frac{40}{143}$
• B. $\frac{70}{143}$
• C. $\frac{3}{13}$
• D. $\frac{10}{13}$

Answer 1

The correct option is D

$\frac{10}{13}$

Explanation for the correct option:

Step:1 Calculate the different ways $3$ balls can be selected from the urn.

The total balls in the urn is $8+5=13$.

Now, the different ways $3$ balls can be selected from the urn is

$$\begin{gathered} C_3^{13} \\ =\frac{13!}{3!\times 10!} \\ =\frac{13\times 12\times 11\times 10!}{3\times 2\times 10!} \\ =286 \end{gathered}$$

Step:2 Calculate that $3$ balls drawn at random from the urn are of different colors.

When $3$ balls contain two different colors, then the favorable (outcome is ($2$ red balls and $1$white ball) or ($1$ red ball and $2$ white balls)

$$\begin{gathered} =C_2^8\times C_1^5+C_1^8\times C_2^5 \\ =\frac{8!}{2\times 6!}\times 5+8\times \frac{5!}{2!\times 3!} \\ =\frac{8\times 7\times 6!}{2\times 6!}\times 5+8\times \frac{5\times 4\times 3!}{2\times 3!} \\ =140+80 \\ =220 \end{gathered}$$

Step:3 Calculate the required probability.

Therefore, the probability that balls of both colors are drawn is

$$\begin{gathered} =\frac{220}{286} \\ =\frac{10}{13} \end{gathered}$$

Therefore, Option(D) is the correct answer.