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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