A customer at Cavallaro’s Fruit Stand picks a sample of 3 oranges at random from a crate containing 60 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges?
another method:
Pr(1 or more)=1-Pr(none rotten)
pr(1 or more)=1- 56/60*55/59*54/58
= 0.189947399
Ah, oranges and probabilities, quite the juicy topic! Let's peel back the layers and calculate the probability of picking 1 or more rotten oranges, shall we?
First, let's find the probability of picking no rotten oranges out of the sample. The chance of picking a good orange is (60 - 4) / 60, because there are 60 oranges in total and 4 of them are rotten. So, the probability of not picking any rotten oranges is ((60 - 4) / 60) * ((60 - 5) / (60 - 1)) * ((60 - 6) / (60 - 2)).
Now, to find the probability of picking 1 or more rotten oranges, we just need to subtract the probability of picking no rotten oranges from 1. Easy peasy!
P = 1 - ((60 - 4) / 60) * ((60 - 5) / (60 - 1)) * ((60 - 6) / (60 - 2))
Now, you can plug that into a calculator and let the math do the work. Remember to grab some popcorn while you wait for the answer, as calculations can sometimes take a bit of time!
To find the probability that the sample contains 1 or more rotten oranges, we need to determine the total number of favorable outcomes and the total number of possible outcomes.
Total number of possible outcomes: There are 60 oranges in the crate, and the customer picks a sample of 3 oranges. Therefore, the total number of possible outcomes is given by the combination formula: nCr = (60C3) = (60! / (3! * (60-3)!)).
Total number of favorable outcomes: To find the number of favorable outcomes, we can consider the complementary event, which is the event of picking 0 rotten oranges. There are 56 non-rotten oranges in the crate. Therefore, the number of favorable outcomes is given by the combination formula: nCr = (56C3) = (56! / (3! * (56-3)!)).
Now, we can calculate the probability as the ratio of the number of favorable outcomes to the number of possible outcomes:
P(1 or more rotten oranges) = 1 - P(0 rotten oranges)
P(1 or more rotten oranges) = 1 - (56C3) / (60C3)
To find the actual numerical value, we can evaluate this expression using a calculator or computer software.
"one or more" = 1, 2 or 3
P(1) = 4/60
P(2) = 4/60 * (4-1)/(60-1)
P(3) = 4/60 * 3/59 * 2/58
Either-or probabilities are found by adding the individual probabilities.