Probability of Defective Products in a Random Sample: A Factory Case Study

In a factory producing 100 products, 4 are found to be defective. When randomly selecting 4 products, we analyze the likelihood that at most three of them are good products. This article breaks down the calculations and explanations involved in determining this probability.

Simple Probability Calculation

Consider a factory producing 100 products, with 4 of them being defective. The probability that a component is good is 96 out of 100, or 96%. Therefore, when the first product is drawn, the chance of drawing a good component is 96 out of 100, or 48/50 simplified. Subsequently, the probability for each subsequent draw, given the previous draw was good, is 95/99, 94/98, and 93/97.

Putting it all together, the probability of drawing 4 good products in a row is:

begin{equation*}frac{96}{100} times frac{95}{99} times frac{94}{98} times frac{93}{97} approx 0.3809 end{equation*}

This calculation can be performed step by step. The probability of drawing a good component on the first draw is 96/100. If the first draw is good, the probability of drawing a good component in the second draw is 95/99. Subsequently, if the second draw is also good, then the probability of drawing a good component in the third draw is 94/98. Finally, if the third draw is also successful, the probability of drawing a good component in the fourth draw is 93/97. Multiplying these probabilities together gives approximately 0.3809, which is intuitively reasonable and represents about 38.1%.

Combinatorial Approach

Another way to derive the probability involves the use of combinations. Let X represent the number of good products in the sample of 4. We want to calculate P(X ≤ 3), which is equivalent to 1 - P(X 4).

Using permutations, the probability of selecting 4 good products out of 96 (since 4 are defective, 96 are good) and 4 products altogether from 100 is calculated by:

begin{equation*} P(X 4) frac{binom{96}{4}}{binom{100}{4}} end{equation*}

Subtracting this from 1 gives us the probability that at most 3 products are good:

begin{equation*} P(X leq 3) 1 - frac{binom{96}{4}}{binom{100}{4}} end{equation*}

This combinatorial approach provides a precise method to determine the probability based on the binomial distribution.

Conclusion

From both a practical and a theoretical standpoint, we can conclude that the probability of drawing at most 3 good products when 4 are randomly selected from a batch of 100 products, with 4 being defective, is approximately 0.3809 or 38.09%. This result is obtained both through step-by-step multiplication of probabilities and through a combinatorial approach using combinations. Both methods confirm that the probability of drawing all 4 good products is about 11.47% (1 - 0.3809).