SOLUTION: Twenty percent of trees in a particular forest have a disease, 30% of the trees are too small to be used for lumber, and 40% are too small to be used for lumber or have a disease.

Algebra ->  Probability-and-statistics -> SOLUTION: Twenty percent of trees in a particular forest have a disease, 30% of the trees are too small to be used for lumber, and 40% are too small to be used for lumber or have a disease.       Log On


   



Question 404204: Twenty percent of trees in a particular forest have a disease, 30% of the trees are too small to be used for lumber, and 40% are too small to be used for lumber or have a disease. What percent of the trees are too small to be used for lumber, given that they have a disease?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
P(diseased) = 0.20
P(too small) = 0.30
P(diseased U too small) = 0.40
==> P(too small|diseased) = P(too small AND diseased)/P(diseased)
=[P(too small) + P(diseased) - P(too small U diseased)]/P(diseased)
=[0.30 + 0.20 - 0.40]/0.20 = 0.10/0.20 = 1/2, or 50% of the trees are too small to be used for lumber, given that they have a disease.