Question
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
answer in standard scientific notation
Answers
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%.
(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%.
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