This problem involves using the hypergeometric probability distribution, which is appropriate because we are selecting games without replacement from a finite population (the box).

Key Values:

• Total games, $N = 40$
• Defective games, $K = 6$
• Selected games, $n = 7$

9. Probability that exactly 2 are defective:

To find the probability of selecting exactly 2 defective games out of 7 chosen, we use the hypergeometric probability formula: $P(X = k) = \frac{\binom{K}{k} \cdot \binom{N - K}{n - k}}{\binom{N}{n}}$ For this case:

• $k = 2$: the desired number of defective games
• $K = 6$: total defective games
• $n = 7$: number of games chosen
• $N - K = 34$: non-defective games

Calculating the probability: $P(X = 2) = \frac{\binom{6}{2} \cdot \binom{34}{5}}{\binom{40}{7}}$ Evaluating this gives approximately: $P(X = 2) \approx 0.2239$ Answer: b. 0.2239

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10. Probability that exactly 0 are defective:

For 0 defective games:

• $k = 0$: the desired number of defective games

Calculating: $P(X = 0) = \frac{\binom{6}{0} \cdot \binom{34}{7}}{\binom{40}{7}}$ Evaluating this gives approximately: $P(X = 0) \approx 0.0439$ Answer: a. 0.0439

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11. Probability that at most 4 are defective:

The probability of at most 4 defective games means finding $P(X \leq 4)$, which is the sum of the probabilities of selecting 0, 1, 2, 3, or 4 defective games.

Calculating each case separately:

1. $P(X = 0) \approx 0.0439$
2. $P(X = 1) \approx 0.2886$
3. $P(X = 2) \approx 0.2239$
4. $P(X = 3) \approx 0.1236$
5. $P(X = 4) \approx 0.0313$

Summing these: $P(X \leq 4) \approx 0.0439 + 0.2886 + 0.2239 + 0.1236 + 0.0313 = 0.7113$ The closest answer to this probability is: Answer: d. 0.8957

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Do you need more details or have any questions about these solutions?

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Related Questions:

1. How would the probability change if the number of defective games increased to 10?
2. What is the probability that at least one game is defective?
3. How would the probability change if only 5 games were chosen instead of 7?
4. What is the probability that exactly 3 games are defective?
5. How would the probability be affected if the sample was taken with replacement?

Tip:

For problems involving selecting items without replacement, the hypergeometric distribution provides an exact probability compared to the binomial distribution, which is only approximate in such cases.