Question 14 Marks
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box.
Answer
View full question & answer→Number of white balls in the bag $= 30$
Let the number of black balls in the box be $x.$
$\therefore $ Total number of balls $= x + 30$
P(drawing a black ball) =$\frac{x}{x+30}$
P(drawing a white ball) = $\frac{30}{x+30}$
It is given that, P(drawing a black ball) = $\frac{2}{5} \times P($ drawing a white ball $)$
$\Rightarrow \frac{x}{x+30}=\frac{2}{5} \times \frac{30}{x+30}$
$\Rightarrow \frac{x}{x+30}=\frac{12}{x+30}$
$\Rightarrow x^2+30 x=12 x+360$
$\Rightarrow x^2+18 x-360=0$
$\Rightarrow x^2+30 x-12 x-360=0$
$\Rightarrow x(x+30)-12(x+30)=0$
$\Rightarrow(x+30)(x-12)=0$
$\Rightarrow x=-30$ or $x=12$
Since number of balls cannot be negative, we reject x = -30
$\Rightarrow x=12$
Therefore, number of black balls in the box is 12.
Let the number of black balls in the box be $x.$
$\therefore $ Total number of balls $= x + 30$
P(drawing a black ball) =$\frac{x}{x+30}$
P(drawing a white ball) = $\frac{30}{x+30}$
It is given that, P(drawing a black ball) = $\frac{2}{5} \times P($ drawing a white ball $)$
$\Rightarrow \frac{x}{x+30}=\frac{2}{5} \times \frac{30}{x+30}$
$\Rightarrow \frac{x}{x+30}=\frac{12}{x+30}$
$\Rightarrow x^2+30 x=12 x+360$
$\Rightarrow x^2+18 x-360=0$
$\Rightarrow x^2+30 x-12 x-360=0$
$\Rightarrow x(x+30)-12(x+30)=0$
$\Rightarrow(x+30)(x-12)=0$
$\Rightarrow x=-30$ or $x=12$
Since number of balls cannot be negative, we reject x = -30
$\Rightarrow x=12$
Therefore, number of black balls in the box is 12.