A customer from Cavallaro's Fruit Stand picks a sample of 3 oranges at random from a crate containing 65 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)
probability of none rotten ... [(65 - 4) / 65]^3
probability 1 or more rotten ... 1 - probability of none rotten
To find the probability of picking 1 or more rotten oranges, we need to consider two scenarios:
1. Picking exactly 1 rotten orange and 2 good oranges.
2. Picking exactly 2 rotten oranges and 1 good orange.
Let's calculate the probability for each scenario and then add them together.
Scenario 1: Picking exactly 1 rotten orange and 2 good oranges.
First, we need to determine the number of ways this can happen. There are 4 rotten oranges in the crate and we need to choose 1 of them. The remaining 2 oranges must be good, so we need to choose 2 good oranges from the remaining 65 - 4 = 61 good oranges. So, the number of ways to pick 1 rotten orange and 2 good oranges is given by the combination formula:
C(4, 1) * C(61, 2) = 4 * (61 * 60) / (2 * 1) = 73260.
The total number of ways to choose 3 oranges from the crate is given by the combination formula:
C(65, 3) = (65 * 64 * 63) / (3 * 2 * 1) = 41640.
Therefore, the probability of picking exactly 1 rotten orange and 2 good oranges is:
P1 = 73260 / 41640.
Scenario 2: Picking exactly 2 rotten oranges and 1 good orange.
Similarly, the number of ways to pick 2 rotten oranges and 1 good orange is given by the combination formula:
C(4, 2) * C(61, 1) = (4 * 3) * 61 = 732.
Therefore, the probability of picking exactly 2 rotten oranges and 1 good orange is:
P2 = 732 / 41640.
Finally, to find the probability of picking 1 or more rotten oranges, we add the probabilities of the two scenarios:
P = P1 + P2.
Now, let's calculate this probability:
P = (73260 / 41640) + (732 / 41640).
P = 73584 / 41640.
P ≈ 0.076.
Therefore, the probability that the sample contains 1 or more rotten oranges is approximately 0.076.