Question

A manufacturing company noticed that 1% of its products are defective. If a dealer orders 300 items from this company, then the probability that the number of defective items is at most one is:

• A. \( \frac{3}{e^3} \)
• B. \( \frac{5}{e^2} \)
• C. \( \frac{3}{e^2} \)
• D. \( \frac{4}{e^3} \)

Hint:

When \( n \) is large and \( p \) is small, use the Poisson distribution as an approximation to the binomial distribution. The mean \( \lambda = np \) and the Poisson probability formula is \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \).

Answer 1

The Correct Option is D

We are given that 1% of the products are defective, and a dealer orders 300 items from the company. We are asked to find the probability that the number of defective items is at most one. 
Step 1: The number of defective items in the order follows a binomial distribution with parameters \( n = 300 \) and \( p = 0.01 \), as 1% of the items are defective. The probability mass function for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] where \( X \) is the number of defective items, \( n \) is the total number of items, \( p \) is the probability of an item being defective, and \( k \) is the number of defective items.
Step 2: We need to find the probability that at most one defective item is in the order, i.e., \( P(X \leq 1) = P(X = 0) + P(X = 1) \).
Step 3: We use the Poisson approximation for the binomial distribution when \( n \) is large and \( p \) is small. The Poisson distribution has the form: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where \( \lambda = np \) is the mean of the distribution. In this case: \[ \lambda = 300 \times 0.01 = 3 \] Thus, the probability of having 0 defective items is: \[ P(X = 0) = \frac{3^0 e^{-3}}{0!} = e^{-3} \] The probability of having 1 defective item is: \[ P(X = 1) = \frac{3^1 e^{-3}}{1!} = 3e^{-3} \] Step 4: The probability of having at most one defective item is: \[ P(X \leq 1) = P(X = 0) + P(X = 1) = e^{-3} + 3e^{-3} = 4e^{-3} \] Thus, the probability that at most one defective item is drawn is \( \frac{4}{e^3} \).