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# 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 $x$.

Given:

A box contains 12 balls out of which $x$ are black.

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.

To do:

We have to find $x$.

Solution:

Total number of balls $=12$

Number of black balls $=x$

This implies,

The total number of possible outcomes $n=12$.

Total number of favourable outcomes(drawing a black ball) $=x$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that the ball drawn is black $=\frac{x}{12}$

When 6 more black balls are put in the box,

The total number of balls $=12+6=18$

Number of black balls $=x+6$

Probability that the ball drawn is black $=\frac{x+6}{18}$

According to the question,

$2\times \frac{x}{12}=\frac{x+6}{18}$

$ \frac{x}{6}=\frac{x+6}{18}$

$3(x)=x+6$

$3x-x=6$

$x=\frac{6}{2}$

$x=3$

**Therefore, $x=3$.**

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