Question

In a consignment of 15 articles, it is found that 3 are defective. If a sample of 5 articles is chosen at random from it, then the probability of having 2 defective articles is:

• A. \(\frac{256}{625}\)
• B. \(\frac{64}{625}\)
• C. \(\frac{128}{625}\)
• D. \(\frac{512}{625}\)

Hint:

For choosing r items out of n, use the combination formula \binom{n}{r} = \frac{n!}{r!(n-r)!}.

Answer 1

The Correct Option is C

We are given: - Total articles = 15 - Number of defective articles = 3 - Number of non-defective articles = 15 - 3 = 12 - Sample size = 5 articles We need to find the probability of selecting exactly 2 defective articles. 

Step 1: Total Possible Combinations 
The total number of ways to select 5 articles out of 15 is: \[ \text{Total combinations} = \binom{15}{5} \] Calculating this: \[ \binom{15}{5} = \frac{15!}{5!(15-5)!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003 \] 

Step 2: Number of Favorable Outcomes 
To have exactly 2 defective articles: - Select 2 defective articles from 3 defective articles: \[ \binom{3}{2} = 3 \] - Select 3 non-defective articles from 12 non-defective articles: \[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

 Step 3: Probability Calculation 
The probability is: \[ P(\text{2 defective}) = \frac{\binom{3}{2} \times \binom{12}{3}}{\binom{15}{5}} \] \[ P = \frac{3 \times 220}{3003} = \frac{660}{3003} = \frac{128}{625} \] 

Step 4: Final Answer 
\[ \boxed{\frac{128}{625}} \] 

Final Answer: (C) \( \frac{128}{625} \)