There are a total of 59 eggs, and 5 of them are brown. So the probability of selecting a brown egg on the first draw is 5/59. After one brown egg has been selected, there are still 4 brown eggs left out of 58 total eggs. So the probability of selecting another brown egg on the second draw is 4/58. Finally, after two brown eggs have been selected, there are still 3 brown eggs left out of 57 total eggs. So the probability of selecting a third brown egg on the third draw is 3/57. We can multiply these probabilities together to get the overall probability:
(5/59) x (4/58) x (3/57) = 0.000298536782 = 2.985 x 10^-4
Therefore, the probability of selecting three brown eggs is approximately 2.985 x 10^-4 or 0.02985%.
A group of hens lays 59 eggs in a single day. On one particular day, there were 5 brown eggs and 54 white eggs. If three eggs are selected at random, without replacement, what is the probability that all three are brown?
answer in standard scientific notation
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