A bag contains 6 red marbles, 5 blue marbles and 8 green marbles. If three marblesare drawn out of the bag, what is the probability, to the nearest 10th of a percent, that all three marbles drawn will be green?
+9674 Number of ways to pick out 3 green marbles = \(\binom{8}3\)
Number of ways to pick out 3 marbles = \(\binom{6 + 5 + 8}3 = \binom{19}3\).
Probability = \(\dfrac{\binom83}{\binom{19}3} = \dfrac{56}{969}\).
Now you can round that to the nearest 10th of a percent, which is just the nearest 0.001.
Interesting fact: The probability is quite close to \(\dfrac{1}{10\sqrt 3}\).
