"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.
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?
2 answers
another method:
Pr(1 or more)=1-Pr(none rotten)
pr(1 or more)=1- 56/60*55/59*54/58
= 0.189947399
Pr(1 or more)=1-Pr(none rotten)
pr(1 or more)=1- 56/60*55/59*54/58
= 0.189947399