$A$ 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 $x$.
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 $x$.
(3) Total number of balls $= 12$.
Total number of black balls $= x$.
The probability of drawing a black ball is $P_1 = \frac{x}{12}$.
If $6$ more black balls are added,the new total number of balls $= 12 + 6 = 18$.
The new number of black balls $= x + 6$.
The new probability of drawing a black ball is $P_2 = \frac{x+6}{18}$.
According to the problem,$P_2 = 2 \times P_1$.
Therefore,$\frac{x+6}{18} = 2 \times \frac{x}{12}$.
$\frac{x+6}{18} = \frac{x}{6}$.
Multiplying both sides by $18$,we get $x + 6 = 3x$.
$2x = 6$,which gives $x = 3$.
Total number of black balls $= x$.
The probability of drawing a black ball is $P_1 = \frac{x}{12}$.
If $6$ more black balls are added,the new total number of balls $= 12 + 6 = 18$.
The new number of black balls $= x + 6$.
The new probability of drawing a black ball is $P_2 = \frac{x+6}{18}$.
According to the problem,$P_2 = 2 \times P_1$.
Therefore,$\frac{x+6}{18} = 2 \times \frac{x}{12}$.
$\frac{x+6}{18} = \frac{x}{6}$.
Multiplying both sides by $18$,we get $x + 6 = 3x$.
$2x = 6$,which gives $x = 3$.
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