To solve this problem, we will use the hypergeometric distribution, as we are drawing from a finite population without replacement.

Given:

• Total number of cloths, $N = 50$
• Number of defective cloths, $D = 10$
• Number of cloths chosen, $n = 12$
• Number of defective cloths to be chosen, $k = 3$

Hypergeometric Probability Formula:

The probability of selecting exactly $k$ defective items from $n$ chosen items without replacement is given by:

$P(X = k) = \frac{\binom{D}{k} \binom{N - D}{n - k}}{\binom{N}{n}}$

Where:

• $\binom{D}{k}$ is the number of ways to choose $k$ defective cloths from $D$ defective cloths.
• $\binom{N - D}{n - k}$ is the number of ways to choose $n - k$ non-defective cloths from the remaining $N - D$ non-defective cloths.
• $\binom{N}{n}$ is the total number of ways to choose $n$ cloths from $N$ cloths.

Step-by-Step Calculation:

1. Calculate the combinations:

   • $\binom{10}{3}$ is the number of ways to choose 3 defective cloths from 10.
   • $\binom{40}{9}$ is the number of ways to choose 9 non-defective cloths from 40.
   • $\binom{50}{12}$ is the total number of ways to choose 12 cloths from 50.

2. Substitute these values into the formula:

$P(X = 3) = \frac{\binom{10}{3} \binom{40}{9}}{\binom{50}{12}}$

Now, let me calculate these combinations:

$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ $\binom{40}{9} = \frac{40 \times 39 \times 38 \times 37 \times 36 \times 35 \times 34 \times 33 \times 32}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 635,040,840$ $\binom{50}{12} = \frac{50 \times 49 \times 48 \times 47 \times 46 \times 45 \times 44 \times 43 \times 42 \times 41 \times 40 \times 39}{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 15,890,700,800$

3. Now, calculate the probability:

$P(X = 3) = \frac{120 \times 635,040,840}{15,890,700,800}$

$P(X = 3) = \frac{76,204,900,800}{15,890,700,800} \approx 0.479$

Conclusion:

The probability that exactly 3 out of the 12 cloths chosen will be defective is approximately 0.479 or 47.9%.

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Would you like more details on any part of the calculation, or do you have any questions?

Here are 5 related questions:

1. What is the general formula for the hypergeometric distribution?
2. How does the hypergeometric distribution differ from the binomial distribution?
3. What is the probability of selecting exactly 2 defective cloths?
4. How would you approach this problem if we were choosing with replacement?
5. What is the expected number of defective cloths in a sample of 12?

Tip: In problems involving the hypergeometric distribution, ensure you are not confusing it with the binomial distribution, which is used when sampling is done with replacement.