Question

A box contains 3 blue balls, 5 red balls, and 7 green balls. You add one ball of each color (3 balls total) to the box and calculate the probability of drawing, at random a green ball from the box. You continue this until the probability of drawing a green ball at random is 8/21. How many balls need to be added?

Answer 1

got to keep adding to make the denominator a multiple of 21, otherwise it will not work

Answer 2

you have 15, add 6, 2 of which are green. that gives 9 green, 21 balls, so \(\frac{9}{21}\) for the first try. that doesn't work

Answer 3

add another 21, 7 of which are green. that gives 16 green, 42 total \(\frac{16}{42}=\frac{8}{21}\) got it on the second try