Question

drawing conclusion

1. 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 alone, and 20 being damaged by insects alone. what is the probability that a randomly selected tree has both a disease and is being damaged by insects? (see example 3)

Explanation:

Step1: Recall the principle of inclusion - exclusion

For two events \(A\) (having a disease) and \(B\) (being damaged by insects), the formula for \(n(A\cup B)\) is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), where \(n(A\cup B)\) is the number of elements in the union of \(A\) and \(B\), \(n(A)\) is the number of elements in \(A\), \(n(B)\) is the number of elements in \(B\), and \(n(A\cap B)\) is the number of elements in the intersection of \(A\) and \(B\).

Let \(A\) be the event that a tree has a disease and \(B\) be the event that a tree is damaged by insects. We know that \(n(A\cup B) = 34\), \(n(A)-n(A\cap B)=18\) (trees with disease alone), and \(n(B)-n(A\cap B)=20\) (trees with insect damage alone). First, we can express \(n(A)=18 + n(A\cap B)\) and \(n(B)=20 + n(A\cap B)\).

Step2: Substitute into the inclusion - exclusion formula

Substitute \(n(A)\), \(n(B)\) and \(n(A\cup B)\) into \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\):

\(34=(18 + n(A\cap B))+(20 + n(A\cap B))-n(A\cap B)\)

Simplify the right - hand side:

\(34=18 + 20+n(A\cap B)\)

\(34 = 38 + n(A\cap B)\)

Then, solve for \(n(A\cap B)\):

\(n(A\cap B)=34 - 38=- 4\)? Wait, that can't be right. Wait, actually, the number of trees with disease alone is \(n(A)-n(A\cap B) = 18\), so \(n(A)=18 + n(A\cap B)\), the number of trees with insect damage alone is \(n(B)-n(A\cap B)=20\), so \(n(B)=20 + n(A\cap B)\), and \(n(A\cup B)=n(A)+n(B)-n(A\cap B)=(18 + n(A\cap B))+(20 + n(A\cap B))-n(A\cap B)=18 + 20 + n(A\cap B)\)

We know that \(n(A\cup B) = 34\), so:

\(18+20 + n(A\cap B)=34\)

\(38 + n(A\cap B)=34\)

Wait, this gives a negative number, which is impossible. Wait, maybe we misinterpret the problem. The number of trees with disease alone is 18, the number with insect damage alone is 20, and the number with either disease or insect damage is 34. Let \(x=n(A\cap B)\). Then the number of trees with disease is \(18 + x\), the number with insect damage is \(20 + x\), and the number with either is \((18 + x)+(20 + x)-x=38 + x\). But we know that the number with either is 34, so \(38 + x=34\) is wrong. Wait, no, the total number of trees with disease or insect damage is the number with disease alone plus the number with insect damage alone plus the number with both. So \(n(A\cup B)=n(A\setminus B)+n(B\setminus A)+n(A\cap B)\), where \(n(A\setminus B)\) is the number with \(A\) alone, \(n(B\setminus A)\) is the number with \(B\) alone.

So \(34 = 18+20 + n(A\cap B)\)

Then \(n(A\cap B)=34-(18 + 20)=34 - 38=-4\). Wait, this is a contradiction. Wait, maybe the problem has a typo? Wait, no, maybe I made a mistake. Wait, the total number of trees is 40, but we are dealing with the 34 trees that have disease or insect damage. Wait, no, the problem says "34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease alone, and 20 being damaged by insects alone". So according to the formula \(n(A\cup B)=n(A\setminus B)+n(B\setminus A)+n(A\cap B)\), so \(34=18 + 20+n(A\cap B)\), so \(n(A\cap B)=34 - 38=- 4\), which is impossible. Wait, maybe the numbers are 18 (disease alone), 14 (insect alone)? No, the problem says 20. Wait, maybe the correct formula is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), where \(n(A)\) is the number with disease (including those with both), \(n(B)\) is the number with insect damage (including those with both). So if 18 have disease alone, then \(n(A)=18 + x\), 20 have insect damage alone, so \(n(B)=20 + x\), and \(n(A\cup B)=34\). Then \(18 + x+20 + x-x=34\), so \(38 + x=34\), \(x=-4\). This is impossible, which means there is an error in the problem statement? Wait, no, maybe I misread. Wait, the total number of trees is 40, but the 34 are among the 40. Wait, maybe the question is about the probability, so we can think in terms of counts. Let's re - express:

Let \(D\) be the set of trees with disease, \(I\) be the set with insect damage. We know that \(|D\cup I| = 34\), \(|D - I|=18\), \(|I - D| = 20\). We know that \(|D\cup I|=|D - I|+|I - D|+|D\cap I|\). So \(34=18 + 20+|D\cap I|\). Then \(|D\cap I|=34-(18 + 20)=34 - 38=-4\). This is impossible, which means there is a mistake. Wait, maybe the numbers are 14 and 20? Or 18 and 14? Wait, maybe the problem meant that 18 have disease (including both) and 20 have insect damage (including both)? Let's try that. If \(n(D)=18\), \(n(I)=20\), \(n(D\cup I)=34\). Then by inclusion - exclusion, \(n(D\cap I)=n(D)+n(I)-n(D\cup I)=18 + 20-34 = 4\). Ah! Maybe the problem statement has a misphrase. It says "18 trees having a disease alone" but maybe it's "18 trees having a disease (including those with both)"? Let's assume that the problem has a typo, and "18 trees having a disease" (not alone), and "20 trees being damaged by insects" (not alone). Then:

Using \(n(D\cup I)=n(D)+n(I)-n(D\cap I)\), we have \(34 = 18+20 - n(D\cap I)\)? No, \(n(D\cup I)=n(D)+n(I)-n(D\cap I)\) implies \(n(D\cap I)=n(D)+n(I)-n(D\cup I)\). If \(n(D) = 18\), \(n(I)=20\), \(n(D\cup I)=34\), then \(n(D\cap I)=18 + 20-34 = 4\). Then the number of trees with both is 4. Then the probability is \(\frac{4}{40}=\frac{1}{10}=0.1\). Wait, maybe the problem meant "18 trees having a disease (not alone)" and "20 trees being damaged by insects (not alone)". Let's check the original problem again: "34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease alone, and 20 being damaged by insects alone". There is a contradiction in the numbers because \(18 + 20=38>34\), which violates the principle that the number of elements in the union cannot be less than the sum of the numbers in each set minus the intersection. So the only way this makes sense is if there is a mistake in the problem, and the "alone" is a mistake. Assuming that \(|D| = 18\) (total with disease), \(|I| = 20\) (total with insect damage), then \(|D\cap I|=|D|+|I|-|D\cup I|=18 + 20 - 34 = 4\). Then the number of trees with both is 4, and the total number of trees is 40. So the probability is \(\frac{4}{40}=0.1\) or \(\frac{1}{10}\).

Wait, maybe the problem is correct, and we have to consider that the 18 and 20 are the numbers with only disease and only insect damage, and the 34 is the number with either. Then the formula \(|D\cup I|=|D - I|+|I - D|+|D\cap I|\) gives \(34 = 18+20+|D\cap I|\), so \(|D\cap I|=34 - 38=-4\), which is impossible. So there must be a typo. The most probable typo is that the number of trees with disease alone is 14, then \(34=14 + 20+|D\cap I|\), \(|D\cap I| = 0\), no. Or the number with insect damage alone is 14, then \(34=18+14+|D\cap I|\), \(|D\cap I| = 2\). But since the problem is given, maybe we misinterpret. Wait, the total number of trees is 40, but the 34 are the ones with disease or insect damage. Wait, no, the problem says "a group of 40 trees... 34 of the trees have a disease or are being damaged by insects...". So the sample space is 40 trees. Let's use the formula for probability: \(P(D\cap I)=\frac{|D\cap I|}{40}\). We know that \(|D\cup I|=|D|+|I|-|D\cap I|\). Also, \(|D|=|D - I|+|D\cap I| = 18+|D\cap I|\), \(|I|=|I - D|+|D\cap I|=20+|D\cap I|\). So \(|D\cup I|=(18 + |D\cap I|)+(20 + |D\cap I|)-|D\cap I|=38+|D\cap I|\). But \(|D\cup I| = 34\), so \(38+|D\cap I|=34\), \(|D\cap I|=-4\). This is impossible, which means there is an error in the problem. But assuming that the problem meant that \(|D| = 18\) (total with disease, including both) and \(|I| = 20\) (total with insect damage, including both), then \(|D\cap I|=18 + 20-34 = 4\), and the probability is \(\frac{4}{40}=0.1\) or \(\frac{1}{10}\).

Answer:

The probability is \(\frac{1}{10}\) (or \(0.1\))