Question

a recent survey asked 150 mathematics professors about how they invest their money between stocks, bonds, and certificates of deposit. the results of the survey are shown. complete parts (a) and (b). out of the 150 mathematics professors surveyed, the survey revealed that: 111 invested in stocks; 101 invested in bonds; 97 invested in certificates of deposit; 80 invested in stocks and bonds; 83 invested in bonds and certificates of deposit; 82 invested in stocks and certificates of deposit; 5 did not invest in any of the three. (a) what is the probability that a randomly chosen professor invested in stocks and bonds? the probability is \\(\frac{8}{15}\\). (type an integer or a simplified fraction.) (b) what is the probability that a randomly chosen professor invested in bonds and certificates of deposit? the probability is \\(\square\\). (type an integer or a simplified fraction.)

Explanation:

Step1: Find the number of professors who invested in at least one of the three.

Total professors are 150, and 5 did not invest in any. So the number of professors who invested in at least one is \(150 - 5=145\).

Step2: Use the principle of inclusion - exclusion for three sets.

Let \(S\) be the set of professors who invested in stocks, \(B\) in bonds, and \(C\) in certificates of deposit. The formula is \(n(S\cup B\cup C)=n(S)+n(B)+n(C)-n(S\cap B)-n(B\cap C)-n(S\cap C)+n(S\cap B\cap C)\). We know \(n(S\cup B\cup C) = 145\), \(n(S)=111\), \(n(B)=101\), \(n(C)=97\), \(n(S\cap B)=80\), \(n(B\cap C)=83\), \(n(S\cap C)=82\). Plugging in the values:
\[145 = 111+101+97 - 80 - 83 - 82 + n(S\cap B\cap C)\]
First, calculate the sum of the single - set counts: \(111 + 101+97=309\)
Then, calculate the sum of the two - set intersections: \(80 + 83+82 = 245\)
So the equation becomes \(145=309 - 245 + n(S\cap B\cap C)\)
\(309-245 = 64\), so \(145=64 + n(S\cap B\cap C)\)
Then \(n(S\cap B\cap C)=145 - 64 = 81\)

Step3: Find the number of professors who invested in bonds and certificates of deposit only? Wait, no. Wait, the question is about the number of professors who invested in bonds and certificates of deposit (including those who also invested in stocks). Wait, no, the formula for \(n(B\cap C)\) is the number of professors who invested in both bonds and certificates of deposit. But we can also think in terms of probability. The probability is the number of professors who invested in bonds and certificates of deposit divided by the total number of professors (150). But we need to find the number of professors who invested in bonds and certificates of deposit. Wait, we know that \(n(B\cap C)\) is given as 83? Wait, no, wait the problem says "83 invested in bonds and certificates of deposit". Wait, but let's check with the inclusion - exclusion. Wait, no, the problem is asking for the probability that a randomly chosen professor invested in bonds and certificates of deposit. The number of professors who invested in bonds and certificates of deposit is 83? Wait, no, wait the total number of professors who invested in at least one is 145. Wait, maybe I made a mistake. Wait, the formula for inclusion - exclusion: \(n(S\cup B\cup C)=n(S)+n(B)+n(C)-n(S\cap B)-n(B\cap C)-n(S\cap C)+n(S\cap B\cap C)\)

We have \(n(S\cup B\cup C) = 145\), \(n(S)=111\), \(n(B)=101\), \(n(C)=97\), \(n(S\cap B)=80\), \(n(B\cap C)=x\) (what we want to find? No, wait the problem says 83 invested in bonds and certificates of deposit? Wait, no, the original problem says "83 invested in bonds and certificates of deposit". Wait, but let's check the total. Wait, maybe the question is about the number of professors who invested in bonds and certificates of deposit (including those who also invested in stocks). But the probability is the number of professors who invested in bonds and certificates of deposit divided by the total number of professors (150). Wait, but let's verify with the inclusion - exclusion. Wait, we found \(n(S\cap B\cap C)=81\). Then, the number of professors who invested in bonds and certificates of deposit is \(n(B\cap C)=n((B\cap C)\cap S)+n((B\cap C)\cap S^c)\). But we know that \(n(B\cap C)=83\) (from the problem statement: "83 invested in bonds and certificates of deposit"). Wait, maybe I misread. Wait, the problem says: "83 invested in bonds and certificates of deposit". So the number of professors who invested in bonds and certificates of deposit is 83. Then the probability is \(\frac{83}{150}\)? Wait, no, that can't be. Wait, no, the total number of professors who invested in at least one is 145. Wait, 150 - 5 = 145. Let's check the inclusion - exclusion again.
\(n(S\cup B\cup C)=n(S)+n(B)+n(C)-n(S\cap B)-n(B\cap C)-n(S\cap C)+n(S\cap B\cap C)\)
\(145=111 + 101+97-80 - n(B\cap C)-82 + n(S\cap B\cap C)\)
\(145=309-(80 + n(B\cap C)+82)+n(S\cap B\cap C)\)
\(145=309-(162 + n(B\cap C))+n(S\cap B\cap C)\)
\(145=147 - n(B\cap C)+n(S\cap B\cap C)\)
\(n(S\cap B\cap C)-n(B\cap C)=145 - 147=- 2\)
This is a contradiction, which means I misread the problem. Wait, the problem says: "83 invested in bonds and certificates of deposit" – no, wait the original problem: "83 invested in bonds and certificates of deposit; 82 invested in stocks and certificates of deposit; 5 did not invest in any of the three." Wait, let's recalculate the inclusion - exclusion correctly.
Let \(n(S) = 111\), \(n(B)=101\), \(n(C)=97\), \(n(S\cap B)=80\), \(n(B\cap C)=83\), \(n(S\cap C)=82\), \(n(\text{none}) = 5\), so \(n(S\cup B\cup C)=150 - 5 = 145\)
Using the formula \(n(S\cup B\cup C)=n(S)+n(B)+n(C)-n(S\cap B)-n(B\cap C)-n(S\cap C)+n(S\cap B\cap C)\)
Substitute the known values:
\(145=111 + 101+97-80 - 83 - 82 + n(S\cap B\cap C)\)
Calculate the right - hand side:
\(111+101 = 212\); \(212 + 97=309\)
\(80+83 = 163\); \(163+82 = 245\)
So \(309-245=64\)
Then \(145=64 + n(S\cap B\cap C)\)
So \(n(S\cap B\cap C)=145 - 64 = 81\)
Now, the number of professors who invested in bonds and certificates of deposit is \(n(B\cap C)=n((B\cap C)\cap S)+n((B\cap C)\cap S^c)\). We know that \(n((B\cap C)\cap S)=n(S\cap B\cap C)=81\), and \(n(B\cap C)=83\), so \(n((B\cap C)\cap S^c)=83 - 81 = 2\). But the question is about the probability that a professor invested in bonds and certificates of deposit (regardless of stocks), so the number of professors who invested in bonds and certificates of deposit is 83. Wait, but the total number of professors is 150, so the probability is \(\frac{83}{150}\)? But that can't be simplified. Wait, no, maybe the question is about the number of professors who invested in bonds and certificates of deposit only? No, the question says "invested in bonds and certificates of deposit", which usually includes those who also invested in stocks. Wait, but let's check the first part: in part (a), the number of professors who invested in stocks and bonds is 80, and the probability is \(\frac{80}{150}=\frac{8}{15}\) (which matches the given answer). So in part (b), the number of professors who invested in bonds and certificates of deposit is 83, so the probability is \(\frac{83}{150}\)? Wait, but 83 and 150 have no common factors (since 83 is a prime number and 83 does not divide 150). Wait, but maybe I made a mistake in the inclusion - exclusion. Wait, no, the problem states "83 invested in bonds and certificates of deposit", so the number of professors who invested in both bonds and certificates of deposit is 83. So the probability is the number of favorable outcomes (83) divided by the total number of outcomes (150).

Answer:

\(\frac{83}{150}\)