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:
The number of soldering defects follows a Poisson distribution with mean \( \lambda = 3 \).
We are asked to find the probability of getting at least two defects, which means \( X \ge 2 \).
Since the Poisson distribution has infinite possible outcomes, it is easier to work with the complement rule instead of adding an endless series of terms.

Step 2: Key Formula or Approach:
The complement of "at least two" is "zero or one" defect, so
\[ P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \]
Instead of computing \( P(0) \) and \( P(1) \) separately from scratch, we can build \( P(1) \) from \( P(0) \) using the Poisson recurrence relation \( P(k) = P(k-1) \times \frac{\lambda}{k} \).

Step 3: Detailed Explanation:
First find \( P(X = 0) \) using the formula directly with \( \lambda = 3 \):
\[ P(0) = e^{-3} = 0.049787 \]
Now use the recurrence relation to jump from \( P(0) \) to \( P(1) \) without recomputing the exponential term:
\[ P(1) = P(0) \times \frac{\lambda}{1} = 0.049787 \times 3 = 0.149361 \]
Add these two probabilities together:
\[ P(X < 2) = P(0) + P(1) = 0.049787 + 0.149361 = 0.199148 \]
Apply the complement rule to get the required probability:
\[ P(X \ge 2) = 1 - 0.199148 = 0.800852 \]

Final Answer:
Rounding 0.800852 to two decimal places gives the answer. \[ \boxed{P(X \ge 2) \approx 0.80} \]