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
To find the probability of selecting all three brown eggs, we need to calculate the probability of each event step-by-step.
Step 1: Determine the probability of selecting the first brown egg.
Out of the 59 eggs laid, there are 5 brown eggs. So the probability of selecting the first brown egg is 5/59.
Step 2: Determine the probability of selecting the second brown egg.
After selecting the first brown egg, there are now 4 brown eggs left out of the remaining 58 eggs. So the probability of selecting the second brown egg is 4/58.
Step 3: Determine the probability of selecting the third brown egg.
After selecting the second brown egg, there are now 3 brown eggs left out of the remaining 57 eggs. So the probability of selecting the third brown egg is 3/57.
To calculate the overall probability of selecting all three brown eggs, we multiply the probabilities of each step together:
P(all three are brown) = (5/59) * (4/58) * (3/57)
This value can then be expressed in standard scientific notation.
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%.