Question

A box contains $12$balls out of which$x$are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If $6$more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find$x$.

Answer 1

Given that, number of black balls $=x$

and total number of balls$=12$

So, the required probability is $p\left(x\right)=x/12$

When 6 more black balls are put in the box, total number of balls $=12+6=18$ then

Number of black balls $=6+x$

And probability of black ball is $p\left(black\right)=\frac{x+6}{18}$

as per condition

$\frac{x+6}{18}=\frac{2x}{12}$
$$\begin{gathered} x+6=3x \\ 6=3x-x \\ 2x=6 \\ x=3 \end{gathered}$$
Hence, number of black balls$=3$