Answer:
To find the probability that all three eggs selected at random are brown, you can use the concept of conditional probability. You'll need to consider the probability of each egg being brown without replacement.
First, find the probability of selecting the first brown egg:
The probability of selecting a brown egg on the first draw is the number of brown eggs divided by the total number of eggs:
P(1st egg is brown) = (Number of brown eggs) / (Total number of eggs) = 7 / 56
After selecting the first brown egg, there are now 6 brown eggs and 55 total eggs left. Now, find the probability of selecting a second brown egg:
P(2nd egg is brown, given 1st is brown) = (Number of remaining brown eggs) / (Number of remaining eggs) = 6 / 55
After selecting the first two brown eggs, there are 5 brown eggs and 54 total eggs left. Now, find the probability of selecting the third brown egg:
P(3rd egg is brown, given 1st and 2nd are brown) = (Number of remaining brown eggs) / (Number of remaining eggs) = 5 / 54
Now, you can find the overall probability that all three eggs are brown by multiplying these probabilities together since they are independent events (assuming you haven't replaced the eggs after each selection):
P(All three eggs are brown) = P(1st egg is brown) * P(2nd egg is brown, given 1st is brown) * P(3rd egg is brown, given 1st and 2nd are brown)
P(All three eggs are brown) = (7/56) * (6/55) * (5/54)
Now, you can calculate this probability:
P(All three eggs are brown) ≈ (7/56) * (6/55) * (5/54) ≈ 0.0001911
So, the probability that all three eggs selected at random are brown is approximately 0.0001911, or about 0.01911%.
Explanation: