Question

A bag contains an assortment of blue and red balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls. Furthermore, the probability of drawing one ball of each color is six times the probability of drawing two blue balls. The number of red and blue balls in the bag is

• A. $6,3$
• B. $3,6$
• C. $2,7$
• D. None of these

Answer 1

The correct option is A

$6,3$

Explanation for the correct option:

Step-1: Formation of equation with given condition:

Let the number of red be $r$ and blue balls be $b$

Then, the probability of picking two red balls, $P\left(R\right)=\frac{{}^{r}C_2}{{}^{r+b}C_2}$

The probability of picking two blue balls, $P\left(B\right)=\frac{{}^{b}C_2}{{}^{r+b}C_2}$

According to question, $P\left(R\right)=5P\left(B\right)$

$\Rightarrow$ $\frac{{}^{r}C_2}{{}^{r+b}C_2}=5\frac{{}^{b}C_2}{{}^{r+b}C_2}$

$\Rightarrow$ $r(r-1)=5b(b-1)\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}\mathbf{]}$ …(1)

The probability of picking one red and one blue ball, $P\left(RB\right)=\frac{{}^{r}C_1.{}^{b}C_1}{{}^{r+b}C_2}$

According to question $P\left(RB\right)=6P\left(B\right)$

$\Rightarrow$ $\frac{{}^{r}C_1.{}^{b}C_1}{{}^{r+b}C_2}=6\frac{{}^{b}C_2}{{}^{r+b}C_2}$

$\Rightarrow$ $2rb=6b(b-1)\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}\mathbf{]}$ ….(2)

Step- 2: Solution of equtions (1) and (2):

$\Rightarrow$$r(r-1)=5b(b-1)...\left(1\right)$ and $2br=6b(b-1)...\left(2\right)$

Divide equation $\left(1\right)$by equation $\left(2\right)$

$$\begin{gathered} \Rightarrow \frac{r(r-1)}{2br}=\frac{5}{6} \\ \Rightarrow 6r-6=10b \end{gathered}$$

By equation $\left(1\right)$

$$\begin{gathered} \frac{(10b+6)}{6}\frac{10b}{6}=5b(b-1) \\ \Rightarrow (10b+6)=18(b-1) \\ \Rightarrow 24=8b \\ \Rightarrow b=3 \end{gathered}$$

From equation $\left(2\right)$

$2br=6b(b-1)...\left(2\right)$

$$\begin{gathered} \Rightarrow 2\times 3r=6\times 3(3-1)...\left(2\right) \\ \Rightarrow r=6 \end{gathered}$$

$\Rightarrow r=6,b=3$

Hence, option ‘A’ is correct.