An urn contains ten marbles, of which give are green, twoareblue, and three are red. Three marbles are to be drawn fromtheurn, one at a time without replacement. What is theprobabilitythat all three marbles drawn will be green?

Answer:[tex] \frac{5\cdot 4\cdot 3}{10\cdot 9 \cdot 8}\approx 0.083[/tex]Step-by-step explanation:Getting all three marbles of green color only happens if every draw is a green marble. On the first marble draw, the urn has 10 marbles in it, out of which 5 are green. So the probability of drawing a green marble on this first draw is [tex]\frac{5}{10}[/tex]Then, once this has happened, the second draw also needs to be a green marble. At this point in the urn there are only 9 marbles left, and only 4 of them are green. So the probability of drawing a green marble at this point is [tex] \frac{4}{9}[/tex]Afterwards, on the last draw, a green marble also needs to be drawn. At this point there are only 8 marbles left on the urn, and only 3 of them are green. So the probability of drawing a green marble on this last draw is [tex] \frac{3}{8}[/tex]Therefore the probability of drawing all three marbles of green color is[tex] \frac{5}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}\approx 0.083[/tex]