A manufacturing concern emplyoing a large number of workers over a period of time and the average absentee rate is 2 workers per shift then probability that exactly two workers will be absent is

Options

A. `(1)/(e^(2))`
B. `(2)/(e^(2))`
C. `(4)/(e^(2))`
D. `2e^(2)`

Answer

To solve the problem, we will use the Poisson distribution formula. The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space, given a known average rate of occurrence.

### Step-by-Step Solution:

1. **Identify the parameters of the problem**:
- The average absentee rate (λ) is given as 2 workers per shift. Thus, λ = 2.

2. **Recall the Poisson distribution formula**:
- The probability of observing exactly k events (in this case, workers absent) in a Poisson distribution is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
- Here, \(e\) is the base of the natural logarithm, \(λ\) is the average rate (2 in this case), and \(k\) is the number of occurrences we want to find the probability for (which is 2 workers absent).

3. **Substitute the values into the formula**:
- We want to find the probability that exactly 2 workers will be absent, so we set \(k = 2\):
\[ P(X = 2) = \frac{e^{-2} \cdot 2^2}{2!} \]

4. **Calculate \(2^2\) and \(2!\)**:
- \(2^2 = 4\)
- \(2! = 2 \times 1 = 2\)

5. **Substitute these values back into the formula**:
- Now we can rewrite the probability:
\[ P(X = 2) = \frac{e^{-2} \cdot 4}{2} \]

6. **Simplify the expression**:
- Simplifying gives:
\[ P(X = 2) = \frac{4}{2} e^{-2} = 2 e^{-2} \]

7. **Final answer**:
- Therefore, the probability that exactly two workers will be absent is:
\[ P(X = 2) = \frac{2}{e^2} \]