A box contains \(6\) red marbles, \(5\) blue marbles, \(2\) yellow marbles, and \(3\) green marbles. Several orange marbles are added to the box. All the marbles in the box are identical except for colour.
A marble is then randomly selected from the box, and the probability that a blue or green marble is selected is \(\tfrac{2}{7}\).
How many orange marbles were added to the box?
The number of blue and green marbles in the box is \(5+3=8\).
Let \(n\) be the total number of marbles in the box after adding some orange marbles. Since the probability of picking a blue or green marble is \(\tfrac{2}{7}\), we must have \(\tfrac{8}{n}=\tfrac{2}{7}.\)
If we multiply the numerator and denominator of the fraction \(\tfrac{2}{7}\) by \(4\), we obtain \(\tfrac{2}{7} = \tfrac{2\times 4}{7\times 4}=\tfrac{8}{28}\). Therefore, \(\tfrac{8}{n}=\tfrac{8}{28}.\) Since the fractions are equal and the numerators are equal, the denominators must also be equal. It follows that \(n=28.\)
In the beginning, there were \(5+6+3+2=16\) marbles in the box. Since there were 16 marbles in the box and there are now 28 marbles in the box, then \(28 - 16=12\) orange marbles were added to the box.
Therefore, \(12\) orange marbles were added to the box.