A group of 40 trees in a forest are not growing properly. A botanist determines that 34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease and 20 being damaged by insects. What is the probability that a randomly selected tree has both a disease and is being damaged by insects?The probability is %.
Question
A group of 40 trees in a forest are not growing properly. A botanist determines that 34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease and 20 being damaged by insects. What is the probability that a randomly selected tree has both a disease and is being damaged by insects?The probability is %.
Solution
To solve this problem, we first need to find out how many trees have both a disease and are being damaged by insects.
Given that 34 trees have either a disease or are being damaged by insects, and knowing that 18 trees have a disease and 20 are being damaged by insects, we can use the principle of inclusion and exclusion to find out how many trees have both conditions.
The principle of inclusion and exclusion states that for any two sets, the size of their union is the sum of their individual sizes minus the size of their intersection.
In this case, the "union" is the set of trees that have either a disease or are being damaged by insects (34 trees), the "individual sizes" are the number of trees with a disease (18 trees) and the number of trees being damaged by insects (20 trees).
So, we can set up the equation:
34 = 18 + 20 - x
Solving for x gives us x = 18 + 20 - 34 = 4.
So, 4 trees have both a disease and are being damaged by insects.
The probability that a randomly selected tree has both a disease and is being damaged by insects is then the number of trees with both conditions divided by the total number of trees.
So, the probability is 4/40 = 0.1 or 10%.
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