This problem involves hypergeometric probability since the manager is sampling without replacement from a finite population. Here's how we approach the problem:

Definitions

• Population size (N): 48 (total radios in the lot)
• Number of defective items (K): 2 (defective radios in the lot)
• Sample size (n): 4 (radios selected for testing)
• Number of successes (x): The number of defective radios found in the sample.

The hypergeometric probability formula is:

$P(X = x) = \frac{\binom{K}{x} \cdot \binom{N - K}{n - x}}{\binom{N}{n}}$

Where:

• $\binom{a}{b}$ is the number of combinations, calculated as $\frac{a!}{b!(a-b)!}$.

Calculations

(a) Probability of finding 2 defective radios ($x = 2$):

$P(X = 2) = \frac{\binom{2}{2} \cdot \binom{46}{2}}{\binom{48}{4}}$

• $\binom{2}{2} = 1$
• $\binom{46}{2} = \frac{46 \cdot 45}{2} = 1035$
• $\binom{48}{4} = \frac{48 \cdot 47 \cdot 46 \cdot 45}{4 \cdot 3 \cdot 2 \cdot 1} = 194580$

Substitute into the formula: $P(X = 2) = \frac{1 \cdot 1035}{194580} \approx 0.0053$

(b) Probability of finding no defective radios ($x = 0$):

$P(X = 0) = \frac{\binom{2}{0} \cdot \binom{46}{4}}{\binom{48}{4}}$

• $\binom{2}{0} = 1$
• $\binom{46}{4} = \frac{46 \cdot 45 \cdot 44 \cdot 43}{4 \cdot 3 \cdot 2 \cdot 1} = 211876$
• $\binom{48}{4} = 194580$ (already computed)

Substitute into the formula: $P(X = 0) = \frac{1 \cdot 211876}{194580} \approx 0.9722$

Final Answers

(a) The probability of finding 2 defective radios is 0.0053.

(b) The probability of finding no defective radios is 0.9722.

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Would you like further explanations or help with similar problems? Here are some related questions to consider:

1. What is the probability of finding exactly 1 defective radio?
2. What is the probability of finding at least 1 defective radio?
3. How does the probability change if the sample size is increased to 6 radios?
4. What is the expected number of defective radios in a random sample of size 4?
5. How would the calculations change if there were 3 defective radios in the lot instead of 2?

Tip: When solving hypergeometric problems, always verify whether sampling is with or without replacement, as this determines whether to use the hypergeometric or binomial distribution.