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Let's break down the problem step by step.

### Given information:
- 20% of the DVDs have a defective disc.
- 14% have a defective case.
- 10% have both a defective disc and a defective case.

We are asked to find the probability for three cases:

#### (a) A defective disc **or** a defective case.
This involves the union of two events: having a defective disc or a defective case. The formula for the union of two events \( A \) and \( B \) is:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Where:
- \( P(A) \) is the probability of a defective disc (20% or 0.20),
- \( P(B) \) is the probability of a defective case (14% or 0.14),
- \( P(A \cap B) \) is the probability of both (10% or 0.10).

So, the probability is:
\[
P(A \cup B) = 0.20 + 0.14 - 0.10 = 0.24 \quad \text{or} \quad 24\%
\]

#### (b) A good disc **or** a good case.
This asks for the probability that at least one of the disc or the case is good. It is the complement of the case where both the disc and the case are defective (i.e., both are bad). So, the formula is:
\[
P(\text{Good disc or good case}) = 1 - P(\text{Both defective})
\]
From the given data, \( P(\text{Both defective}) = 0.10 \).

Thus, the probability is:
\[
P(\text{Good disc or good case}) = 1 - 0.10 = 0.90 \quad \text{or} \quad 90\%
\]

#### (c) A good disc **and** a good case.
This is asking for the probability that both the disc and the case are good. It is the complement of the union of the events where either the disc or the case is defective (i.e., from part (a)). So, the formula is:
\[
P(\text{Good disc and good case}) = 1 - P(\text{Defective disc or defective case})
\]
From part (a), \( P(\text{Defective disc or defective case}) = 0.24 \).

Thus, the probability is:
\[
P(\text{Good disc and good case}) = 1 - 0.24 = 0.76 \quad \text{or} \quad 76\%
\]

### Final answers:
- (a) Probability of a defective disc or a defective case: **24%**
- (b) Probability of a good disc or a good case: **90%**
- (c) Probability of a good disc and a good case: **76%**

Let me know if you'd like further details or explanations! Here are some related questions for further practice:

1. What is the probability that a DVD has only a defective case, but not a defective disc?
2. If you randomly select two DVDs, what is the probability that both have defective discs?
3. What is the probability that a DVD has either a good disc or both defects?
4. How would the probabilities change if 5% more DVDs had defective discs?
5. Can you calculate the conditional probability that a DVD has a defective case given that it has a defective disc?

**Tip:** Remember to visualize the problem using a Venn diagram for more clarity when dealing with overlapping events!

Of all the DVDs in a large shipment, 20% have a defective disc, 14% have a defective case, and 10% have both defects. If you purchase one of the DVDs in this shipment, find the probability that it has the following: (a) a defective disc or a defective case, (b) a good disc or a good case, (c) a good disc and a good case.

Learn how to calculate the probability of defective discs and cases in a DVD shipment. The problem covers key probability concepts, including the inclusion-exclusion principle and the complement rule, with step-by-step solutions for three scenarios.

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