Question

A bag contains 100 articles of which 90 are good and 10 are defective. The articles are tested one by one till all defective are obtained. Which of the following options are correct ?

• A. Probability that first defective article is obtained in second testing is $\frac{1}{11}$
• B. The probability that a particular defective article is obtained in $87^{th}$ testing is $\frac{1}{100}$
• C. Given that second article is defective the probability that first article was good is $\frac{10}{11}$
• D. Given that second article is defective the probability that first article was defective is $\frac{1}{11}$

Answer 1

Answer: (A), (B), (C), (D)

All options (A), (B), (C), and (D) are correct.

Answer 2

Let's check each option:

1. Option (A):
   The event “first defective appears in the 2nd test” requires:

   • 1st test: Good article → $\frac{90}{100}$
   • 2nd test: Defective article → $\frac{10}{99}$

   Total probability:

   $$\frac{90}{100} \times \frac{10}{99} = \frac{900}{9900} = \frac{1}{11}$$

2. Option (B):
   Since the testing order is random, any particular article (defective in case) has an equal chance to appear in any one of the 100 positions. Thus,

   $$P(\text{defective at }87^\text{th}) = \frac{1}{100}$$

3. Option (C):
   Given the 2nd article is defective, let’s calculate:

   • $P(\text{1st good and 2nd defective}) = \frac{90}{100} \times \frac{10}{99}$
   • $P(\text{1st defective and 2nd defective}) = \frac{10}{100} \times \frac{9}{99}$

   Total probability for 2nd defective:

   $$\frac{90 \times 10 + 10 \times 9}{100 \times 99} = \frac{900 + 90}{9900} = \frac{990}{9900} = \frac{1}{10}$$

   Now,

   $$P(\text{1st good} \mid \text{2nd defective}) = \frac{\frac{90}{100} \times \frac{10}{99}}{\frac{1}{10}} = \frac{\frac{900}{9900}}{\frac{1}{10}} = \frac{900}{9900} \times 10 = \frac{9000}{9900} = \frac{10}{11}$$

4. Option (D):
   Complementarily,

   $$P(\text{1st defective} \mid \text{2nd defective}) = 1 - \frac{10}{11} = \frac{1}{11}$$

Thus, all options (A), (B), (C), and (D) are correct.