A group of hens lays 68 eggs in a single day. On one particular day, there were 7 brown eggs and 61 white eggs. If four eggs are selected at random, without replacement, what is the probability that all four are brown?
Give your answer as a decimal or in standard scientific notation with at least 3 significant digits.
Use an asterisk "*" to represent multiplication; do not use the E notation that the calculator uses.
The probability of selecting a brown egg on the first draw is 7/68. Since we are selecting without replacement, the probability of selecting another brown egg on the second draw is 6/67. Similarly, the probability of selecting a third brown egg is 5/66, and the probability of selecting a fourth brown egg is 4/65. Therefore, the probability of selecting four brown eggs is:
(7/68) * (6/67) * (5/66) * (4/65) = 0.000414
Rounding to three significant digits, the answer is 0.000414 or 4.14*10^-4.
To find the probability that all four selected eggs are brown, we need to calculate the probability of selecting a brown egg for each individual selection and multiply these probabilities together.
First, let's find the probability of selecting a brown egg on the first pick. There are a total of 68 eggs, and 7 of them are brown. Therefore, the probability of selecting a brown egg on the first pick is 7/68.
After the first pick, there will be 67 eggs remaining, with 6 brown and 61 white eggs. For the second pick, the probability of selecting a brown egg is 6/67.
Following the same logic for the third and fourth picks, we have:
Probability of selecting a brown egg on the third pick: 5/66
Probability of selecting a brown egg on the fourth pick: 4/65
To find the probability that all four selected eggs are brown, we multiply these probabilities together:
(7/68) * (6/67) * (5/66) * (4/65) = 0.001537 or 1.537*10^(-3)
Therefore, the probability that all four selected eggs are brown is approximately 0.001537 or 1.537*10^(-3).