On a certain day, a foundry manufactured 516 cast aluminum parts. Some of these had major flaws, some had minor flaws, and some had both major and minor flaws. The following table represents the results.

To solve the problem, we first summarize the data in the provided table:

• Major flaw with minor flaw: 18
• Major flaw with no minor flaw: 30
• No major flaw with minor flaws: 80
• No major flaws and no minor flaws: 388

We also know the total number of parts manufactured:

\[ \text{Total parts} = 516 \]

Now we can analyze the data:

1. Parts with major flaws:

   • Major flaw with minor flaw: 18
   • Major flaw with no minor flaw: 30
   • Total with major flaws: \(18 + 30 = 48\)

2. Parts with minor flaws:

   • No major flaw with minor flaws: 80
   • Major flaw with minor flaw: 18
   • Total with minor flaws: \(80 + 18 = 98\)

3. Parts with no flaws:

   • No major flaws and no minor flaws: 388

Now we can proceed to answer the questions:

(a) Probability that a randomly chosen part has a major flaw

The probability \(P(M)\) of a part having a major flaw is given by the number of parts with major flaws divided by the total number of parts:

\[ P(M) = \frac{\text{Number of parts with major flaws}}{\text{Total parts}} = \frac{48}{516} \]

Calculating:

\[ P(M) = \frac{48}{516} \approx 0.093 \]

(b) Probability that a randomly chosen part has a minor flaw

The probability \(P(m)\) of a part having a minor flaw is given by the number of parts with minor flaws divided by the total number of parts:

\[ P(m) = \frac{\text{Number of parts with minor flaws}}{\text{Total parts}} = \frac{98}{516} \]

Calculating:

\[ P(m) = \frac{98}{516} \approx 0.190 \]

(c) Probability that a randomly chosen part has a flaw (major or minor)

To find the probability of a part having at least one flaw (major or minor), we can use the following formula:

\[ P(M \cup m) = P(M) + P(m) - P(M \cap m) \]

Where \(P(M \cap m)\) is the probability of having both major and minor flaws:

\[ P(M \cap m) = \frac{\text{Number of parts with both flaws}}{\text{Total parts}} = \frac{18}{516} \]

Now, let's calculate each probability:

\[ P(M) = \frac{48}{516}, \quad P(m) = \frac{98}{516}, \quad P(M \cap m) = \frac{18}{516} \]

Substituting into the formula:

\[ P(M \cup m) = \frac{48}{516} + \frac{98}{516} - \frac{18}{516} = \frac{48 + 98 - 18}{516} = \frac{128}{516} \]

Calculating:

\[ P(M \cup m) = \frac{128}{516} \approx 0.248 \]

(d) Probability that a randomly chosen part has no major flaw

The probability \(P(\text{no major flaw})\) that a part has no major flaw is the complement of the probability of having a major flaw:

\[ P(\text{no major flaw}) = 1 - P(M) \]

Calculating:

\[ P(\text{no major flaw}) = 1 - \frac{48}{516} = \frac{516 - 48}{516} = \frac{468}{516} \]

Thus:

\[ P(\text{no major flaw}) \approx 0.907 \]

Summary of the Results:

(a) Probability of a major flaw: \(\approx 0.093\)

(b) Probability of a minor flaw: \(\approx 0.190\)

(c) Probability of a flaw (major or minor): \(\approx 0.248\)

(d) Probability of no major flaw: \(\approx 0.907\)