Question

The number of soldering defects that occur in a semiconductor device follows a discrete Poisson distribution with a probability mass function
\[ p(x) = \frac{e^{-\lambda}\lambda^x}{x!} \]
The average number of soldering defects per semiconductor device is 3.
The probability that a randomly selected semiconductor device will have at least two soldering defects is ______ (rounded off to two decimal places).

Hint:

Use the complement rule: P(X≥2) = 1 - P(0) - P(1) for a Poisson distribution with mean 3.

Answer 1

Correct Answer: 0.8

Step 1: Understanding the Concept:
We need to calculate the probability of finding 2 or more defects ($ X \ge 2 $) in a device, given that the distribution is Poisson with an average rate ($ \lambda $) of 3.

Step 2: Key Formula or Approach:
Instead of summing infinite probabilities from 2 to infinity, we use the complementary probability rule:
\[ P(X \ge 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)] \]

Step 3: Detailed Explanation:
Given $ \lambda = 3 $.
Calculate the probability of exactly 0 defects:
\[ P(X = 0) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3} \approx 0.049787 \]
Calculate the probability of exactly 1 defect:
\[ P(X = 1) = \frac{e^{-3} \cdot 3^1}{1!} = 3e^{-3} \approx 0.149361 \]
Sum these probabilities to find $ P(X < 2) $:
\[ P(X < 2) = 0.049787 + 0.149361 = 0.199148 \]
Finally, find the probability of at least two defects:
\[ P(X \ge 2) = 1 - 0.199148 = 0.800852 \]

Final Answer:
Rounded off to two decimal places, the probability comes out to be 0.80. Whenever a question asks for "at least" in a Poisson distribution, calculating the complement is usually the fastest and most practical method. \[ \boxed{P(X \ge 2) \approx 0.80} \]