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$x$.

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Solution

## 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 $=12+6=18$ thenNumber 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}$$x+6=3x\phantom{\rule{0ex}{0ex}}6=3x-x\phantom{\rule{0ex}{0ex}}2x=6\phantom{\rule{0ex}{0ex}}x=3$Hence, number of black balls$=3$

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