A box contains 4 red and 6 green balls. Balls are drawn one at a time at random and NOT replaced, until a red ball appears. What is the probability that exactly three draws are needed?

Exactly three draws are needed when the first two balls are green and the third is red: $(6/10)\times (5/9)\times (4/8)=120/720=1/6$. Because the balls are not replaced the fractions change from draw to draw; keeping them fixed gives $(0.6{)}^2\times 0.4=18/125$, which answers the with-replacement version. The value 2/5 answers yet another question — the chance that the third ball is red whatever the first two are.