Question

A bag contains $b$ blue balls and $r$ red balls. 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 colour is six times the probability of drawing two blue balls. Then which of the following is/are true?

• A. $b+r=9$
• B. $br=18$
• C. $|b-r|=4$
• D. $\frac{b}{r}=2$

Answer 1

Answer: (A), (B)

$br=18$

Given the number of red and blue balls be $r$ and $b$, respectively.
Then, the probability of drawing two red balls is
$p_1=\frac{{}^r{C}_2}{{}^{r+b}{C}_2}=\frac{r(r-1)}{(r+b)(r+b-1)}$
The probability of drawing two blue balls is
$p_2=\frac{{}^b{C}_2}{{}^{r+b}{C}_2}=\frac{b(b-1)}{(r+b)(r+b-1)}$
The probability of drawing one red and one blue ball is
$p_3=\frac{{}^r{C}_1\times {}^b{C}_1}{{}^{r+b}{C}_2}=\frac{2br}{(r+b)(r+b-1)}$
According to question,
$p_1=5p_2$ and $p_3=6p_2$
$\therefore r(r-1)=5b(b-1)$ and $2br=6b(b-1)$
$\Rightarrow r=6,b=3$
$\therefore b+r=9,br=18,|b-r|=3$ and $\frac{b}{r}=\frac{1}{2}$