Probability of Selling the Last Defective Bulb as the 7th Item

In business, particularly in retail, understanding the probability of certain events can help in making informed decisions and managing risks. Consider a scenario where a shopkeeper has 12 light bulbs, out of which 3 are defective. The shopkeeper selects bulbs one at a time at random. What is the probability that the 7th bulb sold is the last defective one?

This problem involves calculating the likelihood of a specific event happening under given conditions. Let's break it down step by step.

Total Bulbs and Problem Condition

There are 12 bulbs in total, with 3 being defective and 9 being non-defective. The 7th bulb must be the last defective one, meaning it is the third defective bulb to be sold.

Steps to Solve

Positioning of Defective Bulbs: For the 7th bulb to be the last defective one, 2 defective bulbs must be sold in the first 6 sales. This leaves the 7th bulb as the last defective one.

Choosing the Positions: We need to select 2 positions from the first 6 for the defective bulbs. The number of ways to choose 2 positions from 6 is given by the combination formula:

[ binom{6}{2} frac{6!}{2!(6-2)!} frac{6 times 5}{2 times 1} 15 ]

Arranging Defective and Non-Defective Bulbs: After selecting the positions for the 2 defective bulbs, the remaining 4 positions in the first 6 must be filled with non-defective bulbs. The last 7th bulb will be the last defective bulb.

Calculating the Total Outcomes: The total number of ways to arrange the 12 bulbs, given that 3 are defective and 9 are non-defective, is calculated as follows:

[ frac{12!}{3! cdot 9!} frac{479001600}{6 cdot 362880} 220 ]

Calculation: The total successful arrangements where the 7th bulb is the last defective one is the product of choosing the positions and arranging them:

[ text{Successful Outcomes} binom{6}{2} cdot 15 15 cdot 15 225 ]

Calculating the Probability: The probability that the 7th bulb sold is the last defective one is:

[ P frac{text{Successful Outcomes}}{text{Total Outcomes}} frac{225}{220} ]

By simplifying this fraction:

[ P frac{225}{220} frac{15}{14} cdot frac{1}{10} frac{3}{44} ]

Final Answer

Therefore, the probability that the 7th bulb sold is the last defective one is:

boxed{frac{3}{44}}

Mastering these calculations can provide valuable insights into supply chain management, inventory control, and decision-making processes in retail and manufacturing environments.