Solution:

Information:

• Scanners Available:
  • 2 electronically defective
  • 1 memory defective
  • 2 in good working order
• Total: 5 scanners
• Objective:
  • Select 2 scanners at random.
  • Compute:
    • (a) Probability of 0 or 1 total defects among the selected.
    • (b) Joint probability distribution of defect counts.

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Part (a): Probability of 0 or 1 Total Defects

We compute probabilities for each scenario using combinatorics.

Total number of ways to choose 2 scanners from 5:

$\binom{5}{2} = \frac{5 \cdot 4}{2} = 10$

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Case 1: 0 Defects

This means both scanners selected are in good working order.

• Number of good scanners: 2.
• Ways to select 2 good scanners: $\binom{2}{2} = 1$
• Probability: $P(0 \text{ defects}) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$

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Case 2: 1 Defect

This means one defective scanner and one good scanner.

• Defective scanners: $2$ electronically defective $+1$ memory defective = $3$.
• Good scanners: $2$.
• Ways to choose 1 defective and 1 good scanner: $\binom{3}{1} \cdot \binom{2}{1} = 3 \cdot 2 = 6$
• Probability: $P(1 \text{ defect}) = \frac{\binom{3}{1} \cdot \binom{2}{1}}{\binom{5}{2}} = \frac{6}{10} = 0.6$

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Total Probability:

$P(0 \text{ or } 1 \text{ defect}) = P(0 \text{ defects}) + P(1 \text{ defect})$ $P(0 \text{ or } 1 \text{ defect}) = \frac{1}{10} + \frac{6}{10} = \frac{7}{10} = 0.7$

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Part (b): Joint Probability Distribution

Let $X$ represent the number of defective scanners in the selection. The possible values of $X$ are $0, 1, 2$.

Case 1: $X = 0$ (No defective scanners)

• Already computed in part (a): $P(X = 0) = \frac{1}{10}$

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Case 2: $X = 1$ (One defective scanner)

• Already computed in part (a): $P(X = 1) = \frac{6}{10} = 0.6$

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Case 3: $X = 2$ (Two defective scanners)

• Ways to select 2 defective scanners from 3 defective: $\binom{3}{2} = 3$
• Probability: $P(X = 2) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10} = 0.3$

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Joint Probability Distribution Table:

$X$ (Number of Defective Scanners)	Probability $P(X)$
0	$0.1$
1	$0.6$
2	$0.3$

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Final Answers:

1. (a) Probability of 0 or 1 total defects: $P(0 \text{ or } 1 \text{ defects}) = 0.7$

2. (b) Joint Probability Distribution:

   • $P(X = 0) = 0.1$
   • $P(X = 1) = 0.6$
   • $P(X = 2) = 0.3$

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Would you like more details or have any questions?
Here are 5 related questions for exploration:

1. How does changing the total number of scanners affect probabilities?
2. What if there were more good scanners? How would the probabilities change?
3. How do probabilities change if selection is not random?
4. Can this problem be solved using a probability tree?
5. What is the expected number of defective scanners in a random selection?

Tip: When solving problems with combinatorics, always carefully count outcomes for each case to ensure completeness.